783. Minimum Distance Between BST Nodes BST節點之間的最小距離
阿新 • • 發佈：2018-02-13
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來自為知筆記(Wiz)
Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.
Example :
Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.
The given tree [4,2,6,1,3,null,null] is represented by the following diagram:
4
/ 2 6
/ \
1 3
while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
Note:
- The size of the BST will be between 2 and
100. - The BST is always valid, each node‘s value is an integer, and each node‘s value is different.
| 12345678910111213141516171819202122232425262728 | # 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): """ :type root: TreeNode :rtype: int """ res = float(‘inf‘) pre = -float(‘inf‘) def find(root): nonlocal res nonlocal pre if root.left: find(root.left) res = min(res, root.val - pre) pre = root.val if root.right: find(root.right) find(root) return res |
來自為知筆記(Wiz)
783. Minimum Distance Between BST Nodes BST節點之間的最小距離