671. 二叉树中第二小的节点

Author: w2xi

README.md

@@ -155,6 +155,7 @@
 |654|[最大二叉树](https://leetcode.cn/problems/maximum-binary-tree/)|[JavaScript](./algorithms/maximum-binary-tree.js)|Medium|
 |657|[机器人能否返回原点](https://leetcode.cn/problems/robot-return-to-origin/)|[JavaScript](./algorithms/robot-return-to-origin.js)|Easy|
 |669|[修剪二叉搜索树](https://leetcode.cn/problems/trim-a-binary-search-tree/)|[JavaScript](./algorithms/trim-a-binary-search-tree.js)|Medium|
+|671|[二叉树中第二小的节点](https://leetcode.cn/problems/second-minimum-node-in-a-binary-tree/)|[JavaScript](./algorithms/second-minimum-node-in-a-binary-tree.js)|Easy|
 |682|[棒球比赛](https://leetcode.cn/problems/baseball-game/)|[JavaScript](./algorithms/baseball-game.js)|Easy|
 |687|[最长同值路径](https://leetcode.cn/problems/longest-univalue-path/)|[JavaScript](./algorithms/longest-univalue-path.js)|Medium|
 |690|[员工的重要性](https://leetcode.cn/problems/employee-importance/)|[JavaScript](./algorithms/employee-importance.js)|Medium|

algorithms/second-minimum-node-in-a-binary-tree.js

@@ -0,0 +1,69 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * 671. 二叉树中第二小的节点
+ * @param {TreeNode} root
+ * @return {number}
+ */
+ var findSecondMinimumValue = function(root) {
+  // 哈希集合去重
+  const set = new Set();
+  const dfs = (root) => {
+      if (root) {
+          set.add(root.val);
+          dfs(root.left);
+          dfs(root.right);
+      }
+  }
+  dfs(root);
+
+  if (set.size < 2) return -1;
+
+  // 两个变量 & 一次遍历 找到次小值
+
+  let first = Number.MAX_VALUE;  // 第一小值
+  let second = Number.MAX_VALUE; // 第二小值
+  // 1,3,-1,2
+  set.forEach(val => {
+      if (val <= first) {
+          second = first;
+          first = val;
+      } else if (val <= second) {
+          second = val;
+      }
+  })
+  return second;
+};
+
+var findSecondMinimumValue2 = function(root) {
+  // 递归
+
+  // root.val = min(root.left.val, root.right.val)
+  // 由于这个条件总是成立, 因此: 根节点必然是全局最小值
+
+  let result = -1;
+
+  const dfs = (root, curr) => {
+    if (root) {
+      if (root.val !== curr) {
+        if (result === -1) { // 第一次赋值
+          result = root.val;
+        } else {
+          result = Math.min(result, root.val);
+        }
+        return;
+      }
+      dfs(root.left, curr);
+      dfs(root.right, curr);
+    }
+  }
+  dfs(root, root.val);
+
+  return result;
+}