LeetCode 530. Minimum Absolute Difference in BST （二叉搜索树中最小绝对差）

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

Example:

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.

 

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题目标签：Binary Search Tree

　　这道题目给了我们一个二叉搜索树，其特性为 左 < 根 < 右。让我们找到树中最小的绝对差值，可以存在任意两点中。如果看到二叉搜索树，一定要条件反射性的想起用 inOrder traverse，所有的值是从小到大的排序。这样就很容易找到最小的绝对差了，对于每一个点，和之前那个点比较一下，遍历完树，就可以找到最小的差值。

 

Java Solution:

Runtime beats 77.33% 

完成日期：07/10/2017

关键词：Binary Search Tree

关键点：利用 inOrder traverse 遍历树，所有点的值排序为从小到大

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution 
{
    int minDiff = Integer.MAX_VALUE;
    TreeNode preNode = null;
    
    public int getMinimumDifference(TreeNode root) 
    {
        inOrder(root);
        return minDiff;
    }
    
    public void inOrder(TreeNode node)
    {
        if(node == null)
            return;
        
        inOrder(node.left);
        
        // get the diff between preNode and node
        if(preNode != null) // because the first time preNode is null
            minDiff = Math.min(minDiff, Math.abs(node.val - preNode.val));
        
        preNode = node;
        
        inOrder(node.right);
    }
}

参考资料：

http://www.cnblogs.com/grandyang/p/6540165.html

 

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