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Leetcode 110. Balanced Binary Tree

2024-08-28 01:58:44   216人阅读

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.

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

这里的方法使用了递归，每个subtree会多次计算level。虽然可以通过OJ, 但是可以使用DP提高效率。 建立一个Hash table, root作为key, level函数的值作为value。
 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]


 


 
Leetcode 110. Balanced Binary Tree






























        

            
                
                  
                
                

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