Leetcode 270. Closest Binary Search Tree Value

Posted 2020-08-26 lettuan

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

Note:

• Given target value is a floating point.
• You are guaranteed to have only one unique value in the BST that is closest to the target.

沿着root向下找即可，每次判段一下是在root左边还是右边。注意没有child的情况。

 

 1 # Definition for a binary tree node.
 2 # class TreeNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution(object):
 9     def closestValue(self, root, target):
10         """
11         :type root: TreeNode
12         :type target: float
13         :rtype: int
14         """
15         #if target == root.val:
16         #    return root.val
17         
18         if target < root.val:
19             if not root.left:
20                 return root.val
21             else:
22                 c1 = self.closestValue(root.left, target)
23                 if abs(c1 - target) < abs(root.val - target):
24                     return c1
25                 else:
26                     return root.val
27         else:
28             if not root.right:
29                 return root.val
30             else:
31                 c2 = self.closestValue(root.right, target)
32                 if abs(c2 - target) < abs(root.val - target):
33                     return c2
34                 else:
35                     return root.val