[LeetCode] 530. Minimum Absolute Difference in BST
作者:互联网
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.
- This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
二叉搜索树的最小绝对差。题目即是题意,找到BST里面任意两点差值的最小值。既然是BST那么还是中序遍历,然后找到中序遍历过程中每两个邻居节点val差值的最小值。我这里提供迭代和递归两种实现。两种实现的复杂度是一样的。
时间O(n)
空间O(n)
Java实现 - 迭代
1 class Solution { 2 public int getMinimumDifference(TreeNode root) { 3 int res = Integer.MAX_VALUE; 4 int prev = Integer.MAX_VALUE; 5 Stack<TreeNode> stack = new Stack<>(); 6 TreeNode cur = root; 7 while (cur != null || !stack.isEmpty()) { 8 while (cur != null) { 9 stack.push(cur); 10 cur = cur.left; 11 } 12 cur = stack.pop(); 13 if (prev != Integer.MAX_VALUE) { 14 res = Math.min(res, cur.val - prev); 15 } 16 prev = cur.val; 17 cur = cur.right; 18 } 19 return res; 20 } 21 }
Java实现 - 递归
1 class Solution { 2 int min = Integer.MAX_VALUE; 3 TreeNode prev; 4 5 public int getMinimumDifference(TreeNode root) { 6 helper(root); 7 return min; 8 } 9 10 public void helper(TreeNode root) { 11 if (root == null) { 12 return; 13 } 14 helper(root.left); 15 if (prev != null) { 16 min = Math.min(min, root.val - prev.val); 17 } 18 prev = root; 19 helper(root.right); 20 } 21 }
相关题目
94. Binary Tree Inorder Traversal
783. Minimum Distance Between BST Nodes - 与本题一模一样
标签:cur,val,min,BST,Minimum,530,prev,root 来源: https://www.cnblogs.com/cnoodle/p/13800662.html