标签:算法
问题描述:
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.
代码:
递归方法
public class Symmetric_Tree { //java public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } //recursively public boolean isSymmetric(TreeNode root) { if(root == null) return true; return isSymmetricTree(root.left, root.right); } public boolean isSymmetricTree(TreeNode left, TreeNode right){ if(left == null && right == null) return true; if(left == null || right == null ||left.val != right.val) return false; boolean lr = isSymmetricTree(left.left, right.right); boolean rl = isSymmetricTree(left.right, right.left); return lr && rl; } }
标签:算法
原文地址:http://blog.csdn.net/chenlei0630/article/details/40514465