小白笔记---------------------------------leetcode(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 [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.

这道题开拓思路，将这颗树复制成两份，然后递归左右比较，思路是看解答才有的，还在想着一次性解决问题，思路堵塞了。。。

/** * 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) {        boolean flag = false;        flag = isMirror(root,root);        return flag;    }    public boolean isMirror(TreeNode root1,TreeNode root2){        if(root1 == null && root2 == null){            return true;        }        if(root1 == null || root2 == null){            return false;        }        return (root1.val == root2.val)             && isMirror(root1.left,root2.right)             && isMirror(root1.right,root2.left);    }}

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