给定一个所有节点为非负值的二叉搜索树,求树中任意两节点的差的绝对值的最小值。
输入: 1 \ 3 / 2 输出: 1 解释: 最小绝对差为1,其中 2 和 1 的差的绝对值为 1(或者 2 和 3)。
注意: 树中至少有2个节点。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def getMinimumDifference(self, root: TreeNode) -> int:
nodes = []
curr = root
prev = float("-inf")
min_diff = float("+inf")
while nodes or curr:
while curr:
nodes.append(curr)
curr = curr.left
curr = nodes.pop()
min_diff = min(min_diff, curr.val - prev)
prev = curr.val
curr = curr.right
return min_diff