Minimum Distance Between BST Nodes
Love problems like this one: https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/
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.
public class Solution { public int MinDiffInBST(TreeNode root) { long previousVal = Int64.MinValue; long minDiff = Int64.MaxValue; _MinDiffInBST(root, ref previousVal, ref minDiff); return (int)minDiff; } public void _MinDiffInBST(TreeNode node, ref long previousVal, ref long minDiff) { if (node == null) return; _MinDiffInBST(node.left, ref previousVal, ref minDiff); minDiff = (previousVal != Int64.MinValue) ? Math.Min(minDiff, node.val - previousVal) : minDiff; previousVal = node.val; _MinDiffInBST(node.right, ref previousVal, ref minDiff); } }
Very nice! Another great thing about Python for lazy people like me is that I never have to think about longs :)
ReplyDeleteclass Solution:
def minDiffInBST(self, root):
state = [None, 10 ** 9]
def helper(node):
if not node:
return
helper(node.left)
prev, min_diff = state
if prev:
min_diff = min(min_diff, node.val - prev)
state[0], state[1] = node.val, min_diff
helper(node.right)
helper(root)
return state[1]