Leetcode 110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

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思路： 先写一个函数treeDepth(root)。递归判段左右子树是否Balanced，如果都是，他们的高度差是不是不超过2。

 1 class Solution(object):
 2     def isBalanced(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: bool
 6         """
 7         if not root:
 8             return True
 9         factor = abs(self.treeDepth(root.left) - self.treeDepth(root.right)) 
10         return factor < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
11         
12     def treeDepth(self, root):
13         if not root:
14             return 0
15         
16         return max(self.treeDepth(root.left), self.treeDepth(root.right)) +

上面这个算法在求treeDepth时，每个节点都会被visit多次。比如算一个node的father node高度时，grand father node高度时，cur node的高度其实都要算一次。可以使用Hash table将已经visited过的node存起来。这样可以保证每个node只求一次高度。

 1 d = {}
 2 class Solution(object):
 3     def isBalanced(self, root):
 4         """
 5         :type root: TreeNode
 6         :rtype: bool
 7         """
 8         if not root:
 9             return True
10         factor = abs(self.level(root.left) - self.level(root.right)) 
11         return factor < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
12         
13     def level(self, root):
14         if not root:
15             return 0
16         
17         if root in d:
18             return d[root]
19         else:
20             d[root] = max(self.level(root.left), self.level(root.right)) + 1
21             return d[root]

 

posted @ 2017-01-25 05:26  lettuan  阅读(112)  评论(0)  编辑  收藏  举报