39平衡二叉樹判斷python
阿新 • • 發佈:2019-01-22
題目:輸入一棵二叉樹,判斷該二叉樹是否是平衡二叉樹。若左右子樹深度差不超過1則為一顆平衡二叉樹。
思路:1、使用獲取二叉樹深度的方法來獲取左右子樹的深度
2、左右深度相減,若大於1返回False
3、通過遞迴對每個節點進行判斷,若全部均未返回False,則返回True
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution :
def getDeepth(self, Root):
if Root is None:
return 0
nright = self.getDeepth(Root.right)
nleft = self.getDeepth(Root.left)
return max(nright, nleft)+1
def IsBalanced_Solution(self, pRoot):
# write code here
if pRoot is None:
return True
right = self.getDeepth(pRoot.right)
left = self.getDeepth(pRoot.left)
if abs(right - left) >1:
return False
return self.IsBalanced_Solution(pRoot.right) and self.IsBalanced_Solution(pRoot.left)
進階思路:
採用後續遍歷,當遍歷到根節點時,每個節點只會遍歷一次
以下程式無法執行,僅提供思路
class Solution:
def IsBalanced(self, pRoot, depth):
if pRoot is None:
return True
if self.IsBalanced(pRoot.right, right) and self.IsBalanced(pRoot.left, left):
diff = abs(left - right)
if diff <= 1:
depth = max(left, right) +1
return True
return False
def IsBalanced_Solution(self, pRoot):
# write code here
depth = 0
return self.IsBalanced(pRoot, depth)