Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3] Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3] Output: false
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
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？
| Language | Runtime | Memory | Submission Time |
|---|---|---|---|
| javascript | 72 ms | 35 MB | 2020/03/23 22:54 |
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isMirror = function(root1, root2) {
if (!root1 && !root2) {
return true;
}
if (!root1 || !root2) {
return false;
}
return (root1.val === root2.val) && isMirror(root1.left, root2.right) && isMirror(root1.right, root2.left);
}
var isSymmetric = function(root) {
return isMirror(root, root);
};
依然使用递归方法，布尔值迭代判断可以写成return 迭代判断条件 && 递归左孩子 && 递归右孩子