标签：style blog color io ar for strong sp div
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.
首先1. top-down 还是 bottom-up. 选择top-down
2. 用递归. 是否需要内嵌小递归. 需要. 其实就是判定root的是否一样的小递归.
3. 逻辑运用小递归. left.left,right.right left.right,right.left
/** * Definition for binary tree * 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 isSymmetric(root.left,root.right); } public boolean isSymmetric(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 isSymmetric(left.left,right.right) && isSymmetric(left.right,right.left); } }
标签：style blog color io ar for strong sp div
原文地址：http://www.cnblogs.com/leetcode/p/4003795.html
