Description

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:

BST

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

Inorder traversal, time O(n), space O(n)

/**
 * 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> list = new ArrayList<>();
        inorder(root, list);
        int minDiff = Integer.MAX_VALUE;
        
        for (int i = 1; i < list.size(); ++i) {
            minDiff = Math.min(minDiff, list.get(i) - list.get(i - 1));
        }
        
        return minDiff;
    }
    
    public void inorder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
}

DFS

自己随便写的，竟然也可以过。

/**
 * 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) {
        if (root == null) {
            return Integer.MAX_VALUE;
        }
        
        int minDiff = Math.min(root.left != null 
                               ? root.val - getMax(root.left) : Integer.MAX_VALUE
                               , root.right != null 
                               ? getMin(root.right) - root.val : Integer.MAX_VALUE);
        minDiff = Math.min(minDiff
                           , Math.min(minDiffInBST(root.left), minDiffInBST(root.right)));
        return minDiff;
    }
    
    public int getMax(TreeNode root) {
        if (root.right == null) {
            return root.val;
        }
        
        return getMax(root.right);
    }
    
    public int getMin(TreeNode root) {
        if (root.left == null) {
            return root.val;
        }
        
        return getMin(root.left);
    }
}