# 101.Symmetric-Tree

## 题目地址

https://leetcode.com/problems/swap-nodes-in-pairs/

## 题目描述

``````Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

1
/ \
2   2
\   \
3    3

Note:
Bonus points if you could solve it both recursively and iteratively.``````

## 代码

### Approach #1 Recursive

Time complexity & Space complexity: O(n)

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
return isMirror(root, root);
}

public boolean isMirror(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null)        return true;
if (t1 == null || t2 == null)        return false;
return (t1.val == t2.val)
&& isMirror(t1.right, t2.left)
&& isMirror(t1.left, t2.right);
}
}``````

### Approach #2 Iterative

``````class Solution {
public boolean isSymetric(TreeNode root) {

while (!queue.isEmpty()) {
TreeNode t1 = queue.poll();
TreeNode t2 = queue.poll();
if (t1 == null && t2 == null)    continue;
if (t1 == null || t2 == null)    return false;
if (t1.val != t2.val)    return false;