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.

 def isBalanced(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         if not root:
             return True
         factor = abs(self.level(root.left) - self.level(root.right))
         return factor < 2 and self.isBalanced(root.right) and self.isBalanced(root.left)

 def level(self, root):
         if not root:
             return 0
         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。

 d = {}
 class Solution(object):
     def isBalanced(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
         if not root:
             return True
         factor = abs(self.level(root.left) - self.level(root.right))
         return factor < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)

     def level(self, root):
         if not root:
             return 0

         if root in d:
             return d[root]
         else:
             d[root] = max(self.level(root.left), self.level(root.right)) + 1
             return d[root]