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530. 二叉搜索树的最小绝对差

English Version

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

解法

中序遍历二叉搜索树，获取当前节点与上个节点的差值的最小值即可。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def getMinimumDifference(self, root: TreeNode) -> int:
        def inorder(root):
            if not root:
                return
            inorder(root.left)
            if self.pre is not None:
                self.min_diff = min(self.min_diff, abs(root.val - self.pre))
            self.pre = root.val
            inorder(root.right)

        self.pre = None
        self.min_diff = 10 ** 5
        inorder(root)
        return self.min_diff

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    private int minDiff = Integer.MAX_VALUE;
    private Integer pre;

    public int getMinimumDifference(TreeNode root) {
        inorder(root);
        return minDiff;
    }

    private void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        if (pre != null) minDiff = Math.min(minDiff, Math.abs(root.val - pre));
        pre = root.val;
        inorder(root.right);
    }
}

Go

var res int
var preNode *TreeNode
func getMinimumDifference(root *TreeNode) int {
    res = int(^uint(0) >> 1)
    preNode = nil
    helper(root)
    return res
}

func helper(root *TreeNode)  {
    if root == nil {
        return
    }
    helper(root.Left)
    if preNode != nil {
        res = getMinInt(res, root.Val - preNode.Val)
    }
    preNode = root
    helper(root.Right)
}

func getMinInt(a,b int) int {
    if a < b {
        return a
    }
    return b
}