[DFS]110.|[DFS]110. Balanced Binary Tree [DFS]110.BalancedBinaryTree

分类：Tree/DFS
时间复杂度: O(nlogn)/O(n)

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.

【[DFS]110.|[DFS]110. Balanced Binary Tree】Example 1:
Given the following tree [3,9,20,null,null,15,7]:

3 / \ 920 /\ 157

Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:

1 / \ 22 / \ 33 / \ 44

Return false.
代码：
O(nlogn)方法:

# Definition for a binary tree node. # class TreeNode: #def __init__(self, x): #self.val = x #self.left = None #self.right = Noneclass Solution: def isBalanced(self, root): """ :type root: TreeNode :rtype: bool """ if root==None: return Trueh_left=self.height(root.left) h_right=self.height(root.right) return abs(h_left-h_right)<=1 and self.isBalanced(root.left) and self.isBalanced(root.right)def height(self, root): if root==None: return 0 else: return max(self.height(root.left),self.height(root.right))+1

O(n)方法:

# Definition for a binary tree node. # class TreeNode: #def __init__(self, x): #self.val = x #self.left = None #self.right = Noneclass Solution: def isBalanced(self, root): """ :type root: TreeNode :rtype: bool """ if root==None: return True def height(root): if root==None: return 0 h_l=height(root.left) h_r=height(root.right) if h_l==-1 or h_r==-1 or abs(h_l-h_r)>1: return -1 else: return max(h_l,h_r)+1 return height(root)!=-1

讨论： 1.经典平衡2叉树，用DFS解（也是递归orz）
2.第一种O（nlogn）的方法好解释
3.第二种O（n）的方法是直接检测高度，当高度差>1的时候直接返回-1，直接返回-1到最终，速度快。