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.

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dfs算法

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