Minimum absolute difference of binary search tree (use medium order traversal as an ordered array)

530. The minimum absolute difference of binary search tree

The difficulty is simple

Give you a binary search tree root node root , return The minimum difference between the values of any two different nodes in the tree .

The difference is a positive number , Its value is equal to the absolute value of the difference between the two values .

Ideas

The problem needs to find the minimum of the absolute value of the difference between any two nodes on the binary search tree .

Notice the binary search tree , Binary search trees are ordered ！

Traverse the binary search tree , What you get is actually an ordered array , Find the minimum difference on an ordered array , Is it very simple ？

package cn.edu.xjtu.carlWay.tree.minAbsoluteDifferenceInBST;

import cn.edu.xjtu.Util.TreeNode.TreeNode;

/** * 530.  The minimum absolute difference of binary search tree  *  Give you a binary search tree root node  root , return   The minimum difference between the values of any two different nodes in the tree  . * <p> *  The difference is a positive number , Its value is equal to the absolute value of the difference between the two values . * <p> * https://leetcode-cn.com/problems/minimum-absolute-difference-in-bst/ */
public class Solution {
    
    int result = Integer.MAX_VALUE;
    TreeNode pre = null;

    /** *  If you think of it as a mid order traversal, it is an ordered array  * * @param root * @return */
    public int getMinimumDifference(TreeNode root) {
    
        if (root == null) return 0;
        traversal(root);
        return result;
    }

    private void traversal(TreeNode root) {
    
        if (root == null) {
    
            return;
        }
        //  Traverse left subtree 
        traversal(root.left);
        
        //  Data processing in the middle 
        if (pre != null) {
    
            result = Math.min(result, root.val - pre.val);
        }
        pre = root;
        
        //  Traversal right subtree 
        traversal(root.right);
    }
}