LeetCode-101.Symmetric Tree

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.

使用队列，两两加入队列，两两出队列

 1 public boolean isSymmetric(TreeNode root) {//树 my
 2         if(null==root){
 3             return true;
 4         }
 5         LinkedList<TreeNode> queue = new LinkedList<>();
 6         queue.add(root.left);
 7         queue.add(root.right);
 8         while(queue.size()!=0){
 9             TreeNode n1 = queue.poll();
10             TreeNode n2 = queue.poll();
11             if(n1==null&&null==n2) continue;
12             if(null==n1||null==n2) return false;
13             if(n1.val!=n2.val) return false;
14             queue.add(n1.left);
15             queue.add(n2.right);
16             queue.add(n1.right);
17             queue.add(n2.left);
18         }
19         return true;
20     }

递归

 1 class Solution {
 2     public boolean isSymmetric(TreeNode root) {//树 my
 3         if(null==root){
 4             return true;
 5         }
 6         return help(root.left,root.right);
 7     }
 8     private boolean help(TreeNode n1 ,TreeNode n2){
 9         if(n1==null&&null==n2) return true;
10         if(null==n1||null==n2) return false;
11         if(n1.val!=n2.val) return false;
12         return help(n1.left,n2.right)&&help(n1.right,n2.left);
13     }
14 }

 

posted @ 2019-04-09 19:45  月半榨菜  阅读(80)  评论(0)  编辑  收藏  举报