1、問題描述

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.

2、邊界條件：root==null

3、思路：1）遞歸，對稱結構是鏡像；左手和右手的關系。所以應該左子樹和右子樹比較，右子樹和左子樹比較是否相同same。但是首先，從root開始分為左右子樹。

base case：左右都為空--true，左右有一個為空--false

2）叠代：一層一層的比較，可以利用隊列，前後取值比較是否相同。

4、代碼實現

1）遞歸

/**
 * 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) {
        if (root == null) {
            return true;
        }
        return isSameTree(root.left, root.right);
    }

    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            
 
return true;
        }
        if (p == null || q == null) {
            return false;
        }

        return p.val == q.val
               && isSameTree(p.left, q.right)
               && isSameTree(p.right, q.left);
    }
}

2）叠代

5、api

leetcode--101. Symmetric Tree