[LeetCode] 110. Balanced Binary Tree_Easy tag: DFS

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.

最简单的思路就是建一个height function, 可以计算每个node的height, 然后abs(left_height - right_height) <2, 再recursive 判断即可. 

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1. Constraints

1) None => True

2. Ideas

DFS      T: O(n^2)   optimal O(n) S; O(n)

     

3. Code

1) T: O(n^2)

class Solution:
    def isBalance(self, root):
        if not root: return True
        def height(root):
            if not root: return 0
            return 1 + max(height(root.left) , height(root.right))
        return abs(height(root.left) - height(root.right)) < 2 and self.isBalance(root.left) and self.isBalance(root.right)

2) T: O(n)  S; O(n)

bottom to up, 一旦发现不符合的,就不遍历, 直接返回-1一直到root, 所以不需要每次来计算node的height. 

class Solution:
    def isBalance(self, root):
        def height(root):
            if not root: return 0
            l, r = height(root.left), height(root.right)
            if -1 in [l, r] or abs(l-r) >1: return -1
            return 1 + max(l,r)
        return height(root) != -1