LeetCode 530 Minimum Absolute Difference in BST

题意

给予一颗非负二叉搜索树, 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

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 getMinimumDifference(TreeNode root) {
        if (root.left != null) {
            getMinimumDifference(root.left);
        }

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

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

Runtime: 1 ms, faster than 95.95% of Java online submissions for Minimum Absolute Difference in BST.
Memory Usage: 38.4 MB, less than 97.37% of Java online submissions for Minimum Absolute Difference in BST.