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530. 二叉搜索树的最小绝对差

原题链接:LeetCode 530. 二叉搜索树的最小绝对差

题目描述

给你一个二叉搜索树的根节点 root ,返回 树中任意两不同节点值之间的最小差值

差值是一个正数,其数值等于两值之差的绝对值。

示例 1:

**输入:**root = [4,2,6,1,3] **输出:**1

示例 2:

**输入:**root = [1,0,48,null,null,12,49] **输出:**1

提示:

树中节点的数目范围是 [2, 104]
0 <= Node.val <= 105

**注意:**本题与 783 https://leetcode.cn/problems/minimum-distance-between-bst-nodes/ 相同

难度: Easy


题解代码

javascript
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var getMinimumDifference = function(root) {
  const vals = dfs(root)
  let min = Infinity
  for(let i = 1; i < vals.length; i++) {
    const diff = Math.abs(vals[i] - vals[i - 1])
    if (diff < min) min = diff
  }
  return min
};

function dfs (root, ret = []) {
  if (!root) return
  if (root.left) dfs(root.left, ret)
  ret.push(root.val)
  if (root.right) dfs(root.right, ret)
  return ret
}

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