LeetCode -- Symmetric Tree

標籤：

Question:

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.

 

Analysis:

問題描述：給出一棵二叉樹，判斷它是否是自己的鏡像樹。即圍繞著中心節點對稱。

思路一：遍曆整棵二叉樹，然後判斷每層的節點是否是對稱的。

思路二：遞迴判斷。首先判斷該節點的左右節點是否對稱，然後儲存左右節點，依次遞迴判斷左節點的左節點與右節點的右節點是否對稱，以及左節點的右節點與右節點的左節點是否對稱。

 

Answer：

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

 

LeetCode -- Symmetric Tree