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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.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1 / 2 3 / 4 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.思路:递归,判断对称的两个节点是否相同。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { private boolean isSameNode(TreeNode left, TreeNode right) { if (left == null && right == null) return true; if (left == null || right == null) return false; return (left.val == right.val) && isSameNode(left.left, right.right) && isSameNode(left.right, right.left); } public boolean isSymmetric(TreeNode root) { if (root == null) return true; return isSameNode(root.left, root.right); } }
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原文地址:http://blog.csdn.net/u011345136/article/details/45310301