101. 对称二叉树(简单题)

思路很简单：递归，如果左子树的左孩子等于右子树的右孩子，并且左子树的右孩子等于右子树的左孩子，即可往下递归。

一开始我居然想用中序遍历。。。。nnd。


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

    public boolean zhongxu(TreeNode left,TreeNode right){

        if(left != null && right != null){
            if(left.val == right.val){
                return zhongxu(left.left,right.right) && zhongxu(left.right,right.left);
            }
        }
        if(left == null && right == null){
            return true;
        }
        return false;
        
    }
}