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feat(two-sum-iv-input-is-a-bst): 递归遍历 + 哈希表 | 双指针
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‎.vscode/settings.json

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{
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"commentTranslate.hover.enabled": true
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"commentTranslate.hover.enabled": true,
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"cSpell.words": ["inorder"]
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}

‎assets/data/category.json

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"path": "../../problemset/minimum-index-sum-of-two-lists/README.md",
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"difficulty": "简单"
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},
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{
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"id": 653,
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"title": "两数之和 IV - 输入 BST",
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"path": "../../problemset/two-sum-iv-input-is-a-bst/README.md",
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"difficulty": "简单"
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},
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{
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"id": 720,
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"title": "词典中最长的单词",
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"path": "../../problemset/happy-number/README.md",
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"difficulty": "简单"
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},
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{
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"id": 653,
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"title": "两数之和 IV - 输入 BST",
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"path": "../../problemset/two-sum-iv-input-is-a-bst/README.md",
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"difficulty": "简单"
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},
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{
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"id": 917,
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"title": "仅仅反转字母",
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"path": "../../problemset/construct-string-from-binary-tree/README.md",
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"difficulty": "简单"
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},
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{
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"id": 653,
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"title": "两数之和 IV - 输入 BST",
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"path": "../../problemset/two-sum-iv-input-is-a-bst/README.md",
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"difficulty": "简单"
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},
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{
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"id": 838,
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"title": "推多米诺",
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"title": "根据二叉树创建字符串",
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"path": "../../problemset/construct-string-from-binary-tree/README.md",
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"difficulty": "简单"
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},
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{
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"id": 653,
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"title": "两数之和 IV - 输入 BST",
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"path": "../../problemset/two-sum-iv-input-is-a-bst/README.md",
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"difficulty": "简单"
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}
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]
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},

‎assets/data/problems.json

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"url": "https://leetcode-cn.com/problems/construct-string-from-binary-tree/",
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"path": "../../problemset/construct-string-from-binary-tree/README.md"
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},
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{
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"id": 653,
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"title": {
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"cn": "两数之和 IV - 输入 BST",
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"en": "two-sum-iv-input-is-a-bst"
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},
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"difficulty": "简单",
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"url": "https://leetcode-cn.com/problems/two-sum-iv-input-is-a-bst/",
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"path": "../../problemset/two-sum-iv-input-is-a-bst/README.md"
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},
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{
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"id": 688,
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"title": {

‎assets/docs/CATEGORY.md

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| 202. [快乐数](../../problemset/happy-number/README.md) | 简单 |
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| 432. [全 O(1) 的数据结构](../../problemset/all-oone-data-structure/README.md) | 困难 |
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| 599. [两个列表的最小索引总和](../../problemset/minimum-index-sum-of-two-lists/README.md) | 简单 |
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| 653. [两数之和 IV - 输入 BST](../../problemset/two-sum-iv-input-is-a-bst/README.md) | 简单 |
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| 720. [词典中最长的单词](../../problemset/longest-word-in-dictionary/README.md) | 简单 |
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| 884. [两句话中的不常见单词](../../problemset/uncommon-words-from-two-sentences/README.md) | 简单 |
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| 1001. [网格照明](../../problemset/grid-illumination/README.md) | 困难 |
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| 165. [比较版本号](../../problemset/compare-version-numbers/README.md) | 中等 |
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| 167. [两数之和 II - 输入有序数组](../../problemset/two-sum-ii-input-array-is-sorted/README.md) | 中等 |
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| 202. [快乐数](../../problemset/happy-number/README.md) | 简单 |
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| 653. [两数之和 IV - 输入 BST](../../problemset/two-sum-iv-input-is-a-bst/README.md) | 简单 |
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| 917. [仅仅反转字母](../../problemset/reverse-only-letters/README.md) | 简单 |
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## 模拟
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| 589. [N 叉树的前序遍历](../../problemset/n-ary-tree-preorder-traversal/README.md) | 简单 |
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| 590. [N 叉树的后序遍历](../../problemset/n-ary-tree-postorder-traversal/README.md) | 简单 |
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| 606. [根据二叉树创建字符串](../../problemset/construct-string-from-binary-tree/README.md) | 简单 |
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| 653. [两数之和 IV - 输入 BST](../../problemset/two-sum-iv-input-is-a-bst/README.md) | 简单 |
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| 838. [推多米诺](../../problemset/push-dominoes/README.md) | 中等 |
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| 1020. [飞地的数量](../../problemset/number-of-enclaves/README.md) | 中等 |
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| 1765. [地图中的最高点](../../problemset/map-of-highest-peak/README.md) | 中等 |
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| 199. [二叉树的右视图](../../problemset/binary-tree-right-side-view/README.md) | 中等 |
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| 589. [N 叉树的前序遍历](../../problemset/n-ary-tree-preorder-traversal/README.md) | 简单 |
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| 606. [根据二叉树创建字符串](../../problemset/construct-string-from-binary-tree/README.md) | 简单 |
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| 653. [两数之和 IV - 输入 BST](../../problemset/two-sum-iv-input-is-a-bst/README.md) | 简单 |
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## 枚举
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‎assets/docs/PROBLEMS.md

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[606. 根据二叉树创建字符串](../../problemset/construct-string-from-binary-tree/README.md)
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[653. 两数之和 IV - 输入 BST](../../problemset/two-sum-iv-input-is-a-bst/README.md)
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[688. 骑士在棋盘上的概率](../../problemset/knight-probability-in-chessboard/README.md)
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[720. 词典中最长的单词](../../problemset/longest-word-in-dictionary/README.md)

‎problemset/two-sum-iv-input-is-a-bst/README.md

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# 两数之和 IV - 输入 BST
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> 难度：简单
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>
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> https://leetcode-cn.com/problems/two-sum-iv-input-is-a-bst/
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## 题目
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给定一个二叉搜索树 `root` 和一个目标结果 `k`，如果 `BST` 中存在两个元素且它们的
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和等于给定的目标结果，则返回 `true`。
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### 示例
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#### 示例 1：
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![two-sum-iv-input-is-a-bst-1](https://user-images.githubusercontent.com/54696834/159190975-736a58d3-a89a-4119-bfb6-cea13459f94b.jpg)
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```
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输入: root = [5,3,6,2,4,null,7], k = 9
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输出: true
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```
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#### 示例 2：
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![two-sum-iv-input-is-a-bst-2](https://user-images.githubusercontent.com/54696834/159190978-0f9da24e-bc9a-47ed-b916-63124d498a11.jpg)
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```
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输入: root = [5,3,6,2,4,null,7], k = 28
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输出: false
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```
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## 解题
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### 广度优先搜索 + 哈希表
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```typescript
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/**
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* 广度优先搜索 + 哈希表
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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const set = new Set<number>();
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const stack = [root];
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while (stack.length > 0) {
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const node = stack.pop()!;
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const diff = k - node.val;
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if (set.has(diff)) return true;
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set.add(node.val);
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node.left && stack.push(node.left);
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node.right && stack.push(node.right);
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}
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return false;
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}
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```
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### 深度优先搜索 + 哈希表
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```typescript
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/**
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* 深度优先搜索 + 哈希表
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget2(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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return dfs(root, k);
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function dfs(node: TreeNode, k: number, set = new Set<number>()): boolean {
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if (set.has(k - node.val)) return true;
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set.add(node.val);
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if (
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(node.left && dfs(node.left, k, set)) ||
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(node.right && dfs(node.right, k, set))
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) {
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return true;
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}
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return false;
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}
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}
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```
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### 深度优先搜索 + 中序遍历 + 双指针
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```typescript
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/**
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* 深度优先搜索 + 中序遍历 + 双指针
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget3(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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const list: number[] = [];
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inorderTraversal(root, list);
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let left = 0;
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let right = list.length - 1;
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while (left < right) {
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const sum = list[left] + list[right];
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if (sum === k) return true;
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if (sum < k) left++;
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else right--;
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}
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return false;
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function inorderTraversal(node: TreeNode | null, list: number[]) {
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if (!node) return;
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inorderTraversal(node.left, list);
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list.push(node.val);
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inorderTraversal(node.right, list);
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}
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}
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```

‎problemset/two-sum-iv-input-is-a-bst/index.spec.ts

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import { TreeNode, createTreeNode } from '~/utils/treeNode';
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import { findTarget, findTarget2, findTarget3 } from '.';
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describe('两数之和 IV - 输入 BST', () => {
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describe('广度优先搜索 + 哈希表', () => {
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testCase(findTarget);
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});
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describe('广度优先搜索 + 哈希表', () => {
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testCase(findTarget2);
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});
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describe('广度优先搜索 + 哈希表', () => {
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testCase(findTarget3);
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});
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});
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function testCase(fn: (root: TreeNode | null, k: number) => boolean) {
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it('示例一', () => {
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const root = createTreeNode([5, 3, 6, 2, 4, null, 7]);
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const k = 9;
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const expected = true;
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expect(fn(root, k)).toBe(expected);
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});
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it('示例二', () => {
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const root = createTreeNode([5, 3, 6, 2, 4, null, 7]);
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const k = 28;
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const expected = false;
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expect(fn(root, k)).toBe(expected);
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});
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}

‎problemset/two-sum-iv-input-is-a-bst/index.ts

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import { TreeNode } from '~/utils/treeNode';
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/**
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* 广度优先搜索 + 哈希表
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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const set = new Set<number>();
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const stack = [root];
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while (stack.length > 0) {
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const node = stack.pop()!;
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const diff = k - node.val;
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if (set.has(diff)) return true;
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set.add(node.val);
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node.left && stack.push(node.left);
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node.right && stack.push(node.right);
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}
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return false;
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}
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/**
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* 深度优先搜索 + 哈希表
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget2(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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return dfs(root, k);
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function dfs(node: TreeNode, k: number, set = new Set<number>()): boolean {
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if (set.has(k - node.val)) return true;
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set.add(node.val);
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if (
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(node.left && dfs(node.left, k, set)) ||
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(node.right && dfs(node.right, k, set))
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) {
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return true;
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}
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return false;
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}
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}
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/**
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* 深度优先搜索 + 中序遍历 + 双指针
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* @desc 时间复杂度 O(N) 空间复杂度 O(N)
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* @param root
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* @param k
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* @returns
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*/
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export function findTarget3(root: TreeNode | null, k: number): boolean {
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if (root === null) return false;
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const list: number[] = [];
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inorderTraversal(root, list);
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let left = 0;
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let right = list.length - 1;
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while (left < right) {
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const sum = list[left] + list[right];
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if (sum === k) return true;
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if (sum < k) left++;
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else right--;
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}
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return false;
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function inorderTraversal(node: TreeNode | null, list: number[]) {
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if (!node) return;
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inorderTraversal(node.left, list);
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list.push(node.val);
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inorderTraversal(node.right, list);
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}
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}

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