给定一个二叉树的 根节点 root,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。
示例 1:
输入: [1,2,3,null,5,null,4]
输出: [1,3,4]
示例 2:
输入: [1,null,3]
输出: [1,3]
示例 3:
输入: []
输出: []
/**
* 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)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var rightSideView = function(root) {
const res = []
if(!root) {
return res
}
let queue = [root] // 维护层次遍历的队列
while(queue.length) {
let curLength = queue.length
for(let i = 0; i < curLength; i++) { //遍历当前层,并输入最右边的节点
let head = queue.pop()
if(head.left !== null) {
queue.unshift(head.left) // 往队列左边插入左节点
}
if(head.right !== null) {
queue.unshift(head.right)
}
if(i === curLength - 1) { // 当前层遍历完,存入最后的节点
res.push(head.val)
}
}
}
return res
};
/**
* 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)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var rightSideView = function(root) {
const res = []
if(!root) {
return res
}
function dfs(root, level) {
if(root === null) {
return;
}
if(level === res.length) { // 该层次第一次访问的节点(先右后左),存入res
res.push(root.val)
}
level++
dfs(root.right, level)
dfs(root.left, level)
}
dfs(root, 0)
return res
};