101. 对称二叉树 (opens new window)

Description

Difficulty: 简单

Related Topics: 树 (opens new window), 深度优先搜索 (opens new window), 广度优先搜索 (opens new window), 二叉树 (opens new window)

给你一个二叉树的根节点 root ， 检查它是否轴对称。

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：true

示例 2：

输入：root = [1,2,2,null,3,null,3]
输出：false

提示：

• 树中节点数目在范围 [1, 1000] 内
• -100 <= Node.val <= 100

**进阶：**你可以运用递归和迭代两种方法解决这个问题吗？

Solution

Language: JavaScript

/**
 * 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 {boolean}
 */
// 方法一：递归
var isSymmetric = function(root) {
 const check = (p, q) => {
  if (!p & !q) return true
  if (!p || !q) return false
  return p.val === q.val && check(p.left, q.right) && check(p.right, q.left)
 }
 return check(root, root)
};

var isSymmetric = function(root) {
 if (!root) return true
 const check = (left, right) => {
  if (!left && !right) return true
  if (!left || !right) return false
  return left.val === right.val && check(left.left, right.right) && check(left.right, right.left)
 }
 return check(root.left, root.right)
};