LeetCode in TypeScript
530. Minimum Absolute Difference in BST
Easy
Given the root of a Binary Search Tree (BST), return the minimum absolute difference between the values of any two different nodes in the tree.
Example 1:
Input: root = [4,2,6,1,3]
Output: 1
Example 2:
Input: root = [1,0,48,null,null,12,49]
Output: 1
Constraints:
- The number of nodes in the tree is in the range
[2, 104]. 0 <= Node.val <= 105
Note: This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
Solution
import { TreeNode } from '../../com_github_leetcode/treenode'
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function getMinimumDifference(root: TreeNode | null): number {
let ans = Number.MAX_SAFE_INTEGER
let prev: number | null = null
function dfs(node: TreeNode | null) {
if (!node) {
return
}
dfs(node.left)
if (prev !== null) {
ans = Math.min(ans, Math.abs(node.val - prev))
}
prev = node.val
dfs(node.right)
}
dfs(root)
return ans
}
export { getMinimumDifference }