1. 题目
2. 思路
- 通过后序遍历得到两棵二叉树的叶子节点数组;
- 再判断两个数组是否相同
3. 代码实现
/**
* 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} root1
* @param {TreeNode} root2
* @return {boolean}
*/
var leafSimilar = function(root1, root2) {
let leaf1 = [], leaf2 = []
getLeaf(root1, leaf1)
getLeaf(root2, leaf2)
console.log(leaf1, leaf2)
function getLeaf(node, arr) {
// 空节点
if (node === null) return
// 左右子树
let left = node.left,
right = node.right
getLeaf(left, arr)
getLeaf(right, arr)
// 叶子节点
if (left === null && right === null) {
arr.push(node.val)
}
}
if (leaf1.length !== leaf2.length) return false
return leaf1.toString() === leaf2.toString()
};
4. 参考
代码简洁 一种还不错的解法
JavaScript数组(四):判断数组相等的4种方法