LeetCode 272. Closest Binary Search Tree Value II（二叉搜索树查找）

原题网址：https://leetcode.com/problems/closest-binary-search-tree-value-ii/

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.

思路：二叉搜索树的中序遍历是一个有序数列，假设有一个长度为k的双端队列，我们只需要遍历的时候，把遍历到的节点加入到队尾，如果队列长度超过k，则需要从对头或者队尾删除掉一个。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private LinkedList<Integer> result = new LinkedList<>();
private void traverse(TreeNode root, double target, int k) {
if (root.left != null) traverse(root.left, target, k);
if (result.size() < k) result.add(root.val);
else if (Math.abs(root.val-target) < Math.abs(result.getFirst()-target)) {
result.addLast(root.val);
result.removeFirst();
} else {
return;
}
if (root.right != null) traverse(root.right, target, k);
}
public List<Integer> closestKValues(TreeNode root, double target, int k) {
if (root == null) return result;
traverse(root, target, k);
return result;
}
}

也可以使用两个栈来从小到大、从大到小来排序 ，即从两端向中间收敛，参考：https://leetcode.com/discuss/55240/ac-clean-java-solution-using-two-stacks