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)])

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