给一棵非空二叉搜索树以及一个target值,找到在BST中最接近给定值的节点值
样例1
输入: root = {5,4,9,2,#,8,10} and target = 6.124780
输出: 5
解释:
二叉树 {5,4,9,2,#,8,10},表示如下的树结构:
5
/ 4 9
/ / 2 8 10
样例2
输入: root = {3,2,4,1} and target = 4.142857
输出: 4
解释:
二叉树 {3,2,4,1},表示如下的树结构:
3
/ 2 4
/
1
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: the given BST
@param target: the given target
@return: the value in the BST that is closest to the target
"""
def closestValue(self, root, target):
# write your code here
"""
特别简单好理解的方法,非递归:
如果当前root值比target大,就暂且把这个root值当成上限,然后往左边走
如果当前root值比target小,就暂且把这个root值当成下限,然后往右边走
左右摇摆着走,知道发现两个最接近target的值,由于是inplace的更新上下限,而且不递归,所以没有额外的空间损耗
O(h) time and O(1) space
"""
if not root:
return None
lower, upper = root, root
node = root
while node:
if node.val < target:
lower = node
node = node.right
elif node.val > target:
upper = node
node = node.left
else:
return node.val
return lower.val if abs(upper.val-target) > abs(lower.val-target) else upper.val
原文:https://www.cnblogs.com/bonelee/p/11624355.html