程序媛詹妮弗
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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 [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.

 

题目

给定二叉树,判断其是否左右对称

 

思路

DFS

 

代码

 1 /* 
 2 Time:O(n),
 3 Space: O(logn) 即 O(h) 树的deepest hight
 4 */
 5 public class Solution {
 6     public boolean isSymmetric(TreeNode root) {
 7         if (root == null) return true;
 8         return isSymmetric(root.left, root.right);
 9     }
10     private static boolean isSymmetric(TreeNode p, TreeNode q) {
11         if (p == null && q == null) return true;   // 终止条件
12         if (p == null || q == null) return false;  // 终止条件
13         return p.val == q.val      // 三方合并
14                 && isSymmetric(p.left, q.right)
15                 && isSymmetric(p.right, q.left);
16     }
17 }

 

posted on 2018-10-21 14:59  程序媛詹妮弗  阅读(115)  评论(0编辑  收藏  举报