Skip to content

Commit 4786b6e

Browse files
committed
add LeetCode 199. 二叉树的右视图
1 parent e7d8936 commit 4786b6e

File tree

1 file changed

+115
-0
lines changed

1 file changed

+115
-0
lines changed

Diff for: 二叉树/LeetCode 199. 二叉树的右视图.md

+115
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,115 @@
1+
![](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL2Nob2NvbGF0ZTE5OTkvY2RuL2ltZy8yMDIwMDgyODE0NTUyMS5qcGc?x-oss-process=image/format,png)
2+
>仰望星空的人,不应该被嘲笑
3+
4+
## 题目描述
5+
6+
给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
7+
8+
示例:
9+
10+
```javascript
11+
输入: [1,2,3,null,5,null,4]
12+
输出: [1, 3, 4]
13+
解释:
14+
15+
1 <---
16+
/ \
17+
2 3 <---
18+
\ \
19+
5 4 <---
20+
```
21+
22+
来源:力扣(LeetCode)
23+
链接:https://leetcode-cn.com/problems/binary-tree-right-side-view
24+
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
25+
26+
27+
## 解题思路
28+
`DFS`,每一层只能取一个元素,那么我们搜的时候优先考虑右孩子即可。
29+
![](https://img-blog.csdnimg.cn/20200925163146472.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjQyOTcxOA==,size_16,color_FFFFFF,t_70#pic_center)
30+
参考 <a href="https://leetcode-cn.com/problems/binary-tree-right-side-view/solution/shen-du-you-xian-sou-suo-by-shetia-2/">shetia</a> 大佬图解
31+
32+
```javascript
33+
/**
34+
* Definition for a binary tree node.
35+
* function TreeNode(val) {
36+
* this.val = val;
37+
* this.left = this.right = null;
38+
* }
39+
*/
40+
/**
41+
* @param {TreeNode} root
42+
* @return {number[]}
43+
*/
44+
var rightSideView = function(root) {
45+
if(!root) return [];
46+
let res = [];
47+
let dfs = (step,root) => {
48+
if(res.length === step){
49+
res.push(root.val);
50+
}
51+
// 优先遍历右孩子,再遍历左孩子
52+
root.right && dfs(step+1,root.right);
53+
root.left && dfs(step+1,root.left);
54+
}
55+
dfs(0,root);
56+
return res;
57+
};
58+
```
59+
60+
**解法二**
61+
62+
`BFS`,对于每一层取队列中对首元素,然后放入队列的时候,如果有对应左右孩子的话,优先放右孩子,再放左孩子。
63+
64+
```javascript
65+
/**
66+
* Definition for a binary tree node.
67+
* function TreeNode(val) {
68+
* this.val = val;
69+
* this.left = this.right = null;
70+
* }
71+
*/
72+
/**
73+
* @param {TreeNode} root
74+
* @return {number[]}
75+
*/
76+
var rightSideView = function (root) {
77+
if (!root) return [];
78+
let queue = [root];
79+
let res = [];
80+
while (queue.length) {
81+
let size = queue.length;
82+
res.push(queue[0].val);
83+
while (size--) {
84+
let node = queue.shift();
85+
// 优先放右孩子
86+
node.right && queue.push(node.right);
87+
node.left && queue.push(node.left);
88+
}
89+
}
90+
return res;
91+
};
92+
```
93+
94+
## 最后
95+
文章产出不易,还望各位小伙伴们支持一波!
96+
97+
往期精选:
98+
99+
<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a>
100+
101+
<a href="https://github.com/Chocolate1999/leetcode-javascript">leetcode-javascript:LeetCode 力扣的 JavaScript 解题仓库,前端刷题路线(思维导图)</a>
102+
103+
小伙伴们可以在Issues中提交自己的解题代码,🤝 欢迎Contributing,可打卡刷题,Give a ⭐️ if this project helped you!
104+
105+
106+
<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~
107+
108+
![](https://img-blog.csdnimg.cn/2020090211491121.png#pic_center)
109+
110+
```javascript
111+
学如逆水行舟,不进则退
112+
```
113+
114+
115+

0 commit comments

Comments
 (0)