[LeetCode] 530. Minimum Absolute Difference in BST

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/

二叉搜索树的最小绝对差。

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。

思路是中序遍历。注意对于一个 BST 而言，他的中序遍历的结果是递增的，所以遍历结果里面两个相邻的 node 的差值很好计算。我这里提供迭代和递归两种实现。两种实现的复杂度是一样的。

时间O(n)

空间O(n)

Java实现 - 迭代

 1 class Solution {
 2     public int getMinimumDifference(TreeNode root) {
 3         int res = Integer.MAX_VALUE;
 4         int prev = Integer.MAX_VALUE;
 5         Stack<TreeNode> stack = new Stack<>();
 6         TreeNode cur = root;
 7         while (cur != null || !stack.isEmpty()) {
 8             while (cur != null) {
 9                 stack.push(cur);
10                 cur = cur.left;
11             }
12             cur = stack.pop();
13             if (prev != Integer.MAX_VALUE) {
14                 res = Math.min(res, cur.val - prev);
15             }
16             prev = cur.val;
17             cur = cur.right;
18         }
19         return res;
20     }
21 }

Java实现 - 递归

 1 class Solution {
 2     int min = Integer.MAX_VALUE;
 3     TreeNode prev;
 4 
 5     public int getMinimumDifference(TreeNode root) {
 6         helper(root);
 7         return min;
 8     }
 9 
10     public void helper(TreeNode root) {
11         if (root == null) {
12             return;
13         }
14         helper(root.left);
15         if (prev != null) {
16             min = Math.min(min, root.val - prev.val);
17         }
18         prev = root;
19         helper(root.right);
20     }
21 }

CNoodle

Let the code speak for you.

[LeetCode] 530. Minimum Absolute Difference in BST

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