Formatted question description: https://leetcode.ca/all/783.html

783. Minimum Distance Between BST Nodes (Medium)

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

Example :

Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.

The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

      4
    /   \
  2      6
 / \    
1   3  

while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Note:

  1. The size of the BST will be between 2 and 100.
  2. The BST is always valid, each node’s value is an integer, and each node’s value is different.

Solution 1.

// OJ: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
// Time: O(N)
// Space: O(log(N))
class Solution {
private:
    int ans = INT_MAX;
    TreeNode *prev = NULL;
    void inorder(TreeNode *root) {
        if (!root) return;
        inorder(root->left);
        if (prev) ans = min(ans, root->val - prev->val);
        prev = root;
        inorder(root->right);
    }
public:
    int minDiffInBST(TreeNode* root) {
        inorder(root);
        return ans;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int minDiffInBST(TreeNode root) {
            List<Integer> inorderTraversal = inorderTraversal(root);
            int minDiff = Integer.MAX_VALUE;
            int size = inorderTraversal.size();
            for (int i = 1; i < size; i++) {
                int diff = inorderTraversal.get(i) - inorderTraversal.get(i - 1);
                minDiff = Math.min(minDiff, diff);
            }
            return minDiff;
        }
    
        public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> inorderTraversal = new ArrayList<Integer>();
            Stack<TreeNode> stack = new Stack<TreeNode>();
            TreeNode node = root;
            while (!stack.isEmpty() || node != null) {
                while (node != null) {
                    stack.push(node);
                    node = node.left;
                }
                TreeNode visitNode = stack.pop();
                inorderTraversal.add(visitNode.val);
                node = visitNode.right;
            }
            return inorderTraversal;
        }
    }
    
  • // OJ: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
    // Time: O(N)
    // Space: O(log(N))
    class Solution {
    private:
        int ans = INT_MAX;
        TreeNode *prev = NULL;
        void inorder(TreeNode *root) {
            if (!root) return;
            inorder(root->left);
            if (prev) ans = min(ans, root->val - prev->val);
            prev = root;
            inorder(root->right);
        }
    public:
        int minDiffInBST(TreeNode* root) {
            inorder(root);
            return ans;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def minDiffInBST(self, root):
            """
            :type root: TreeNode
            :rtype: int
            """
            vals = []
            def inOrder(root):
                if not root:
                    return 
                inOrder(root.left)
                vals.append(root.val)
                inOrder(root.right)
            inOrder(root)
            return min([vals[i + 1] - vals[i] for i in xrange(len(vals) - 1)])
    

All Problems

All Solutions