LeetCode OJ 101. Symmetric Tree

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

For example, this binary tree is symmetric:

But the following is not:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

判断一棵树是不是对称的，那么我们需要对比两个位置对称的节点，首先判断这两个节点的值是否相等，然后判断这两个节点的子树是否对称。这就是递归的思路，从根节点的左右子树开始，递归向下。代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSymmetric(TreeNode root) {
12         if(root == null) return true;
13         return isSame(root.left, root.right);
14     }
15     
16     public boolean isSame(TreeNode root1, TreeNode root2){
17         if(root1==null && root2==null) return true;
18         if(root1!=null && root2!=null){
19             if(root1.val != root2.val) return false;
20             else{
21                 boolean a = isSame(root1.left, root2.right);
22                 boolean b = isSame(root1.right, root2.left);
23                 if(a==true && b==true) return true;
24                 else return false;
25             }
26         }
27         return false;
28     }
29 }