LeetCode 783 Minimum Distance Between BST Nodes

题意

给予一颗二叉搜索树, 返回任意两个节点之间的最小相差值.

注: 树至少有两个节点.

例 :

给予树:


   1
    \
     4
    / \
   2   7


返回: 1 (1 和 2 之间相差 1).

解法

这道题很像: Minimum Absolute Difference in BST, 解法甚至可以通用.

因为是一颗二叉搜索树, 所以采用中序遍历可以得到所有值从小到大的排列, 那么将每个节点与上个节点的值 prev 进行比较得出相差值 answer, 判断相差值与上个相差值, 将更小的存起来. 直到遍历完整棵树.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int prev = -1;
    private int answer = Integer.MAX_VALUE;
    public int minDiffInBST(TreeNode root) {
        if (root.left != null) {
            minDiffInBST(root.left);
        }

        if (prev != -1) {
            answer = Math.min(answer, root.val - prev);
        }
        
        prev = root.val;

        if (root.right != null) {
            minDiffInBST(root.right);
        }
        return answer;
    }
}

Runtime: 0 ms, faster than 100.00% of Java online submissions for Minimum Distance Between BST Nodes.
Memory Usage: 33.5 MB, less than 100.00% of Java online submissions for Minimum Distance Between BST Nodes.