Leetcode 110. Balanced Binary Tree

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

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

1 defisbalanced (self, root):2         """3 : Type Root:treenode4 : Rtype:bool5         """6         if  notRoot:7             returnTrue8Factor = ABS (Self.level (Root.left)-Self.level (root.right))9         returnFactor < 2 andself.isbalanced (Root.right) andself.isbalanced (root.left)Ten  One defLevel (self, root): A         if  notRoot: -             return0 -         returnMax (Self.level (Root.left), Self.level (root.right)) + 1

The method here uses recursion, and each subtree calculates the level multiple times. Although it is possible to pass OJ, you can use DP to improve efficiency. Create a hash table, root as key, and the level function as value.

1D = {}2 classsolution (object):3     defisbalanced (self, root):4         """5 : Type Root:treenode6 : Rtype:bool7         """8         if  notRoot:9             returnTrueTenFactor = ABS (Self.level (Root.left)-Self.level (root.right)) One         returnFactor < 2 andself.isbalanced (Root.left) andself.isbalanced (root.right) A          -     defLevel (self, root): -         if  notRoot: the             return0 -          -         ifRootinchD: -             returnD[root] +         Else: -D[root] = max (Self.level (Root.left), Self.level (root.right)) + 1 +             returnD[root]

Leetcode 110. Balanced Binary Tree