题目:
Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.
Note:
- Given target value is a floating point.
- You may assume k is always valid, that is: k ≤ total nodes.
- You are guaranteed to have only one unique set of k values in the BST that are closest to the target.
Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
Hint:
- Consider implement these two helper functions:
getPredecessor(N)
, which returns the next smaller node to N.getSuccessor(N)
, which returns the next larger node to N.
- Try to assume that each node has a parent pointer, it makes the problem much easier.
- Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
- You would need two stacks to track the path in finding predecessor and successor node separately.
链接: http://leetcode.com/problems/closest-binary-search-tree-value-ii/
题解:
一开始思路非常不明确,看了不少discuss也不明白为什么。在午饭时间从头仔细想了一下,像Closest Binary Search Tree Value I一样,追求O(logn)的解法可能比较困难,但O(n)的解法应该不难实现。我们可以使用in-order的原理,从最左边的元素开始,维护一个Deque或者doubly linked list,将这个元素的值从后端加入到Deque中,然后继续遍历下一个元素。当Deque的大小为k时, 比较当前元素和队首元素与target的差来尝试更新deque。循环结束条件是队首元素与target的差更小或者遍历完全部元素。这样的话时间复杂度是O(n), 空间复杂度应该是O(k)。
Time Complexity - O(n), Space Complexity - O(k)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> closestKValues(TreeNode root, double target, int k) { LinkedList<Integer> res = new LinkedList<>(); inOrder(root, target, k, res); return res; } private void inOrder(TreeNode root, double target, int k, LinkedList<Integer> res) { if(root == null) { return; } inOrder(root.left, target, k, res); if(res.size() == k) { if(Math.abs(res.get(0) - target) >= Math.abs(root.val - target)) { res.removeFirst(); res.add(root.val); } else { return; } } else { res.add(root.val); } inOrder(root.right, target, k, res); } }
二刷:
还是使用inorder traversal来遍历树,同时维护一个size为k的LinkedList或者Deque。这个原理类似于sliding window。
Java:
Time Complexity - O(n), Space Complexity - O(k)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> closestKValues(TreeNode root, double target, int k) { LinkedList<Integer> res = new LinkedList<>(); inOrderTraversal(root, target, k, res); return res; } private void inOrderTraversal(TreeNode root, double target, int k, LinkedList<Integer> res) { if (root == null) { return; } inOrderTraversal(root.left, target, k, res); if (res.size() < k) { res.add(root.val); } else if(res.size() == k) { if (Math.abs(res.getFirst() - target) > (Math.abs(root.val - target))) { res.removeFirst(); res.addLast(root.val); } else { return; } } inOrderTraversal(root.right, target, k, res); } }
Reference:
https://leetcode.com/discuss/55261/efficient-python
https://leetcode.com/discuss/70577/java-in-order-traversal-1ms-solution
https://leetcode.com/discuss/64713/clear-java-solution-with-one-stack-one-linkedlist
https://leetcode.com/discuss/64713/clear-java-solution-with-one-stack-one-linkedlist
https://leetcode.com/discuss/69220/2-ms-o-n-and-6-ms-o-logn-java-solution
https://leetcode.com/discuss/55240/ac-clean-java-solution-using-two-stacks
https://leetcode.com/discuss/55682/o-logn-java-solution-with-two-stacks-following-hint
https://leetcode.com/discuss/55486/java-two-stacks-iterative-solution
https://leetcode.com/discuss/64713/clear-java-solution-with-one-stack-one-linkedlist
https://leetcode.com/discuss/71820/java-5ms-iterative-following-hint-o-klogn-time-and-space
https://leetcode.com/discuss/70577/java-in-order-traversal-1ms-solution