272. Closest Binary Search Tree Value II

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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:

1. Consider implement these two helper functions:
   1. getPredecessor(N), which returns the next smaller node to N.
   2. getSuccessor(N), which returns the next larger node to N.
2. Try to assume that each node has a parent pointer, it makes the problem much easier.
3. Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
4. 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);
    }
}

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

272. Closest Binary Search Tree Value II

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原文地址：http://www.cnblogs.com/yrbbest/p/5031304.html