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783. 二叉搜索树节点最小距离

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, 100]
  • 0 <= Node.val <= 105

 

注意：本题与 530：https://leetcode.cn/problems/minimum-absolute-difference-in-bst/ 相同

解法

方法一：中序遍历

中序遍历二叉搜索树，获取当前节点与上个节点差值的最小值即可。

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDiffInBST(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if root is None:
                return
            dfs(root.left)
            nonlocal ans, prev
            ans = min(ans, abs(prev - root.val))
            prev = root.val
            dfs(root.right)

        ans = prev = inf
        dfs(root)
        return ans

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int ans;
    private int prev;
    private int inf = Integer.MAX_VALUE;

    public int minDiffInBST(TreeNode root) {
        ans = inf;
        prev = inf;
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, Math.abs(root.val - prev));
        prev = root.val;
        dfs(root.right);
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    const int inf = INT_MAX;
    int ans;
    int prev;

    int minDiffInBST(TreeNode* root) {
        ans = inf, prev = inf;
        dfs(root);
        return ans;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        ans = min(ans, abs(prev - root->val));
        prev = root->val;
        dfs(root->right);
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDiffInBST(root *TreeNode) int {
    inf := 0x3f3f3f3f
    ans, prev := inf, inf
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        ans = min(ans, abs(prev-root.Val))
        prev = root.Val
        dfs(root.Right)
    }
    dfs(root)
    return ans
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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var minDiffInBST = function (root) {
    let ans = Number.MAX_SAFE_INTEGER,
        prev = Number.MAX_SAFE_INTEGER;
    const dfs = root => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, Math.abs(root.val - prev));
        prev = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
};

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