[DFS]110. Balanced Binary Tree

• 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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

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 = None

class Solution:
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root==None:
            return True
        
        h_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 = None

class 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

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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。

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