783. 二叉搜索树节点最小距离
作者:互联网
给定一个二叉搜索树的根节点 root,返回树中任意两节点的差的最小值。
示例:
输入: root = [4,2,6,1,3,null,null]
输出: 1
解释:
注意,root是树节点对象(TreeNode object),而不是数组。
给定的树 [4,2,6,1,3,null,null] 可表示为下图:
4
/ \
2 6
/ \
1 3
最小的差值是 1, 它是节点1和节点2的差值, 也是节点3和节点2的差值。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/minimum-distance-between-bst-nodes
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def minDiffInBST(self, root: TreeNode) -> int: def preorderTraversal(root): node, output = root, [] while node: if not node.left: output.append(node.val) node = node.right else: predecessor = node.left while predecessor.right and predecessor.right is not node: predecessor = predecessor.right if not predecessor.right: output.append(node.val) predecessor.right = node node = node.left else: predecessor.right = None node = node.right return output nodes=preorderTraversal(root) nodes.sort(reverse=True) _min=float('inf') for i in range(len(nodes)-1): _min=min(_min,nodes[i]-nodes[i+1]) return _min
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def minDiffInBST(self, root: TreeNode) -> int: if not root:return pre=float('-inf') _min=float('inf') stack=[] cur=root while stack or cur: while cur: stack.append(cur) cur=cur.left cur=stack.pop() _min=min(_min,cur.val-pre) pre=cur.val cur=cur.right return _min
标签:node,right,cur,783,self,min,二叉,root,节点 来源: https://www.cnblogs.com/xxxsans/p/13637190.html