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Copy path94.二叉树的中序遍历.js
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94.二叉树的中序遍历.js
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/*
* @lc app=leetcode.cn id=94 lang=javascript
*
* [94] 二叉树的中序遍历
*
* https://leetcode.cn/problems/binary-tree-inorder-traversal/description/
*
* algorithms
* Easy (75.91%)
* Likes: 1481
* Dislikes: 0
* Total Accepted: 873.9K
* Total Submissions: 1.2M
* Testcase Example: '[1,null,2,3]'
*
* 给定一个二叉树的根节点 root ,返回 它的 中序 遍历 。
*
*
*
* 示例 1:
*
*
* 输入:root = [1,null,2,3]
* 输出:[1,3,2]
*
*
* 示例 2:
*
*
* 输入:root = []
* 输出:[]
*
*
* 示例 3:
*
*
* 输入:root = [1]
* 输出:[1]
*
*
*
*
* 提示:
*
*
* 树中节点数目在范围 [0, 100] 内
* -100 <= Node.val <= 100
*
*
*
*
* 进阶: 递归算法很简单,你可以通过迭代算法完成吗?
*
*/
// @lc code=start
/**
* 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 inorderTraversal = function (root) {
// 递归版本
const res = []
const fn = (node) => {
if(!node) return []
fn(node.left)
res.push(node.val)
fn(node.right)
}
fn(root)
return res
}
var inorderTraversal = function (root) {
// 迭代版本
const res = [],
stack = []
let cur = root
while (cur || stack.length) {
while (cur) {
stack.push(cur)
cur = cur.left
}
cur = stack.pop()
res.push(cur.val)
cur = cur.right
}
return res
}
// @lc code=end