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;
};

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