530. Minimum Absolute Difference in BST

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

Example:

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Input:
   1
    \
     3
    /
   2
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.

解法1：

in-order的traversal是一个排序过后的数组，用一个prev记录上一次访问过的node，然后相邻的比较一下。
用一个global variable寄存
C++

1

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int min = Integer.MAX_VALUE;
    TreeNode prev = null;
    public int getMinimumDifference(TreeNode root) {
        // bst in order traversal is a sorted array
        // Record the prev node and current node
        // Calculate the difference and compare with global min
        inorder(root);
        return min;        
    }
    private void inorder(TreeNode root) {
        if (root == null) {
            return;
        }
        inorder(root.left);
        if (prev != null) {
            min = Math.min(min, Math.abs(prev.val - root.val));
        }
        prev = root;
        inorder(root.right);
    }
}