[LeetCode] 530. Minimum Absolute Difference in BST

最新推荐文章于 2025-06-18 21:18:22 发布

DragonWar%

最新推荐文章于 2025-06-18 21:18:22 发布

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原文链接：https://segmentfault.com/a/1190000016997230

本文探讨了在非负值二叉搜索树中寻找任意两节点间最小绝对差值的问题，提供了两种解决方案：迭代使用栈和递归中序遍历。通过详细解析算法思路，帮助读者理解并掌握高效求解此问题的方法。

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Problem

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

   1
    \
     3
    /
   2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Note: There are at least two nodes in this BST.

Solution

Iterative -- using stack

class Solution {    
    public int getMinimumDifference(TreeNode root) {
        int min = Integer.MAX_VALUE;
        if (root == null) return min;
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode pre = null;
        while (!stack.isEmpty() || root != null) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            TreeNode cur = stack.pop();
            if (pre != null) {
                min = Math.min(min, cur.val-pre.val);
            }
            pre = cur;
            root = cur.right;
        }
        return min;
    }
}

recursive -- in-order

class Solution {
    Integer pre = null;
    int min = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return min;
        getMinimumDifference(root.left);

        if (pre != null) {
            min = Math.min(min, root.val-pre);
        }
        pre = root.val;

        getMinimumDifference(root.right);

        return min;
    }
}