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.

Python3JavaC++GoTypeScriptRustJavaScript

# 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

/**
 * 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);
    }
}

/**
 * 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 = [&](this auto&& dfs, TreeNode* root) -> void {
            if (!root) {
                return;
            }
            dfs(root->left);
            ans = min(ans, root->val - pre);
            pre = root->val;
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};

/**
 * 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(root.Right)
    }
    dfs(root)
    return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDiffInBST(root: TreeNode | null): number {
    let [ans, pre] = [Infinity, -Infinity];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn min_diff_in_bst(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        const inf: i32 = 1 << 30;
        let mut ans = inf;
        let mut pre = -inf;

        fn dfs(node: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32, pre: &mut i32) {
            if let Some(n) = node {
                let n = n.borrow();
                dfs(n.left.clone(), ans, pre);
                *ans = (*ans).min(n.val - *pre);
                *pre = n.val;
                dfs(n.right.clone(), ans, pre);
            }
        }

        dfs(root, &mut ans, &mut pre);
        ans
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDiffInBST = function (root) {
    let [ans, pre] = [Infinity, -Infinity];
    const dfs = root => {
        if (!root) {
            return;
        }
        dfs(root.left);
        ans = Math.min(ans, root.val - pre);
        pre = root.val;
        dfs(root.right);
    };
    dfs(root);
    return ans;
};

Comments