783. Minimum Distance Between BST Nodes - Easy

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:

          4
        /         2      6
     / \
    1   3  

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.

M1: inorder traversal (recursive)

time: O(n), space: O(h)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int prev = Integer.MAX_VALUE, minDist = Integer.MAX_VALUE;

    public int minDiffInBST(TreeNode root) {
        if(root == null) {
            return 0;
        }
        helper(root);
        return minDist;
    }

    public void helper(TreeNode node) {
        if(node == null) {
            return;
        }
        helper(node.left);
        if(prev != Integer.MAX_VALUE) {
            minDist = Math.min(minDist, Math.abs(node.val - prev));
        }
        prev = node.val;
        helper(node.right);
    }
}

M2: inorder traversal (iterative)

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) {
        if(root == null) {
            return 0;
        }
        int minDist = Integer.MAX_VALUE;
        LinkedList<TreeNode> s = new LinkedList<>();
        TreeNode cur = root;
        int prev = Integer.MAX_VALUE;
        while(cur != null || !s.isEmpty()) {
            while(cur != null) {
                s.offerFirst(cur);
                cur = cur.left;
            }

            cur = s.pollFirst();
            if(prev != Integer.MAX_VALUE) {
                minDist = Math.min(minDist, Math.abs(prev - cur.val));
            }
            prev = cur.val;
            cur = cur.right;
        }
        return minDist;
    }
}

原文地址：https://www.cnblogs.com/fatttcat/p/10251061.html