101. Symmetric Tree

题目：

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.

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链接：http://leetcode.com/submissions/detail/11328058/

一刷

recursive

 1 def isSymmetric(p, q):
 2     if not p and not q:
 3         return True
 4     if not p or not q:
 5         return False
 6     
 7     return p.val == q.val and isSymmetric(p.left, q.right) and isSymmetric(p.right, q.left)
 8         
 9 class Solution(object):
10     def isSymmetric(self, root):
11         if not root:
12             return True
13         return isSymmetric(root.left, root.right)

iterative，利用tuple来成对比较，答案是抄的。。。

 1 class Solution(object):
 2     def isSymmetric(self, root):
 3         if not root:
 4             return True
 5         stack = [(root.left, root.right)]
 6         
 7         while stack:
 8             l_elem, r_elem = stack.pop()
 9             if not l_elem and not r_elem:
10                 continue
11             if not l_elem or not r_elem:
12                 return False
13             if l_elem.val == r_elem.val:
14                 stack.extend([(l_elem.left, r_elem.right), (l_elem.right, r_elem.left)])
15             else:
16                 return False
17         return True

参考：

https://leetcode.com/discuss/14944/recursively-and-iteratively-solution-in-python

 

2/16/2017, Java

performance不够好，二刷再解决

 1 public class Solution {
 2     public boolean isSymmetric(TreeNode root) {
 3         if (root == null || root.left == null && root.right == null) return true;
 4         return isSymmetric(root.left, root.right);
 5     }
 6     private boolean isSymmetric(TreeNode p, TreeNode q) {
 7         if (p == null && q == null) return true;
 8         if (p == null && q != null || p != null && q == null || p.val != q.val) return false;
 9         if (isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left)) return true;
10         return false;
11     }
12 }

 

posted @ 2016-06-17 10:39  panini  阅读(107)  评论(0)  编辑  收藏  举报