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:

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

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

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：

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

     public boolean isSame(TreeNode root1, TreeNode root2){
         if(root1==null && root2==null) return true;
         if(root1!=null && root2!=null){
             if(root1.val != root2.val) return false;
             else{
                 boolean a = isSame(root1.left, root2.right);
                 boolean b = isSame(root1.right, root2.left);
                 if(a==true && b==true) return true;
                 else return false;
             }
         }
         return false;
     }
 }