530. 二叉搜索树的最小绝对差(简单)

1,问题描述

530. 二叉搜索树的最小绝对差

难度:简单

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:

img

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输入:root = [4,2,6,1,3]
输出:1

示例 2:

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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.com/problems/minimum-distance-between-bst-nodes/ 相同

2,初步思考

​ 解法:搜索二叉树中序遍历时,数组是单调递增的,所以只用求俩俩差值最小就行

3,代码处理

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import support.TreeNode;

import java.util.ArrayList;
import java.util.List;

public class _530二叉搜索树的最小绝对差 {

// 解法:中序遍历优化
// 只用遍历一次即可
public int getMinimumDifference_option(TreeNode root) {
temp = null;
min = Integer.MAX_VALUE;
treeSlipMid_v2(root);
return min;
}

Integer temp = null;

private void treeSlipMid_v2(TreeNode root) {
if (root == null) return;
treeSlipMid_v2(root.left);
if (temp != null) {
min = Math.min(min, Math.abs(root.val - temp));
}
temp = root.val;
treeSlipMid_v2(root.right);
}

// 中序遍历求解
public int getMinimumDifference_mid(TreeNode root) {
list = new ArrayList<>();
treeSlipMid(root);
int min = Integer.MAX_VALUE;
for (int i = 1; i < list.size(); i++) {
min = Math.min(min, list.get(i) - list.get(i - 1));
}
return min;
}

private void treeSlipMid(TreeNode root) {
if (root == null) return;
treeSlipMid(root.left);
list.add(root.val);
treeSlipMid(root.right);
}

List<Integer> list;

// 错误题解
public int getMinimumDifference_fail(TreeNode root) {
if (root != null) {
if (root.left != null) {
min = Math.min(min, -root.left.val + root.val);
getMinimumDifference_fail(root.left);
}
if (root.right != null) {
min = Math.min(min, root.right.val - root.val);
getMinimumDifference_fail(root.right);
}
}
return min;
}

int min = Integer.MAX_VALUE;

public static void main(String[] args) {
_530二叉搜索树的最小绝对差 minDiffInBST = new _530二叉搜索树的最小绝对差();
TreeNode root = new TreeNode(4, new TreeNode(2, new TreeNode(1), new TreeNode(3)), new TreeNode(6));
System.out.println(minDiffInBST.getMinimumDifference_option(root));
}
}