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

Python Code
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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

Java Code
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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);
    }
}

C++ Code
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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);
    }
};

Go Code
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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
}