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783. Minimum Distance Between BST Nodes

Description

Given the root of a Binary Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree.

 

Example 1:

Input: root = [4,2,6,1,3]
Output: 1

Example 2:

Input: root = [1,0,48,null,null,12,49]
Output: 1

 

Constraints:

  • The number of nodes in the tree is in the range [2, 100].
  • 0 <= Node.val <= 105

 

Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/

Solutions

Solution 1: Inorder Traversal

The problem requires us to find the minimum difference between the values of any two nodes. Since the inorder traversal of a binary search tree is an increasing sequence, we only need to find the minimum difference between the values of two adjacent nodes in the inorder traversal.

We can use a recursive method to implement the inorder traversal. During the process, we use a variable $\textit{pre}$ to save the value of the previous node. This way, we can calculate the minimum difference between the values of two adjacent nodes during the traversal.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary search tree.

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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: Optional[TreeNode]):
            if root is None:
                return
            dfs(root.left)
            nonlocal pre, ans
            ans = min(ans, root.val - pre)
            pre = root.val
            dfs(root.right)

        pre = -inf
        ans = 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 final int inf = 1 << 30;
    private int ans = inf;
    private int pre = -inf;

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

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = 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:
    int minDiffInBST(TreeNode* root) {
        const int inf = 1 << 30;
        int ans = inf, pre = -inf;
        auto dfs = [&](auto&& dfs, TreeNode* root) -> void {
            if (!root) {
                return;
            }
            dfs(dfs, root->left);
            ans = min(ans, root->val - pre);
            pre = root->val;
            dfs(dfs, root->right);
        };
        dfs(dfs, root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDiffInBST(root *TreeNode) int {
    const inf int = 1 << 30
    ans, pre := inf, -inf
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        ans = min(ans, root.Val-pre)
        pre = root.Val
        dfs(