Leetcode: Count Complete Tree Nodes

Get idea from 西施豆腐渣.

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Question

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

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Analysis

If write down the answer as wrong solution described below, the time will exceeded. How to cut down its time ? We can get the number before counting the whole tree, by comparing the left height and right height. This is the smart way to save time.

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Solution

wrong answer (Time Limit Exceeded)

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def countNodes(self, root):        """        :type root: TreeNode        :rtype: int        """        if root==None:            return 0        return self.helper(root)    def helper(self, root):        if root==None:            return 0        return 1 + self.helper(root.left) + self.helper(root.right)

My Solution

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def countNodes(self, root):        """        :type root: TreeNode        :rtype: int        """        if root==None:            return 0        lh = self.getlh(root)        rh = self.getrh(root)        if lh==rh:            return 2**lh-1        else:            return 1 + self.countNodes(root.left) + self.countNodes(root.right)    def getlh(self, root):        h = 0        while root!=None:            h += 1            root = root.left        return h    def getrh(self, root):        h = 0        while root!=None:            h += 1            root = root.right        return h