Description

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

tree

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Note: There are at least two nodes in this BST.

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 getMinimumDifference(TreeNode root) {
        List<Integer> list = new LinkedList<>();
        inorderTraversal(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 inorderTraversal(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        
        inorderTraversal(root.left, list);
        list.add(root.val);
        inorderTraversal(root.right, list);
    }
}

Inorder Traversal, time O(n), space O(1)

可以用辅助的成员变量来记录前一个访问过的节点pre，这样可以省去额外的空间开销。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int minDiff = Integer.MAX_VALUE;
    private TreeNode pre = null;
    
    public int getMinimumDifference(TreeNode root) {
        inorderTraversal(root);
        return minDiff;
    }
    
    public void inorderTraversal(TreeNode root) {
        if (root == null) {
            return;
        }
        
        inorderTraversal(root.left);
        if (pre != null) {
            minDiff = Math.min(minDiff, root.val - pre.val);
        }
        pre = root;
        inorderTraversal(root.right);
    }
}