L 272. Closest Binary Search Tree Value II -----------H ~~~~~！！！！！！！！！

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.

Example:

Input: root = [4,2,5,1,3], target = 3.714286, and k = 2

    4
   / \
  2   5
 / \
1   3

Output: [4,3]

Follow up:

Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)? 

A:

首先想到都弄到一个数组里，然后全部排序。  然而不是optimal

思路是找到k个比他小的， 和k个比他大的。 然后前后两头儿检查 删除一个。

O（ k + log N ) time,  O(k) space

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> closestKValues(TreeNode* root, double target, int k) {
        deque<int> res;
        // find up to k values, that smaller than(or equal to) target
        find_k_smallerEq(root, target, k, res);
        deque<int> resBigger;
        find_k_bigger(root, target, k, resBigger);
        for (auto v : resBigger) {
            res.push_back(v);
        }

        while (res.size() > k) {
            if (abs(res.front() - target) >= abs(res.back() - target)) {
                res.pop_front();
            }
            else {
                res.pop_back();
            }
        }
        return vector<int>(res.begin(), res.end());
    }
private:
    void find_k_smallerEq(TreeNode* root, double target, const int k, deque<int>& res) {
        if (not root || res.size() >= k)
            return;
        if (root->val > target) {
            find_k_smallerEq(root->left, target, k, res);
            return;
        }
        find_k_smallerEq(root->right, target, k, res);
        if (res.size() < k)
            res.push_front(root->val);
        find_k_smallerEq(root->left, target, k, res);
    }
    void find_k_bigger(TreeNode* root, double target, const int k, deque<int>& res) {
        if (not root || res.size() >= k) {
            return;
        }
        if (root->val <= target) {
            find_k_bigger(root->right, target, k, res);
            return;
        }
        find_k_bigger(root->left, target, k, res);
        if (res.size() < k)
            res.push_back(root->val);
        find_k_bigger(root->right, target, k, res);
    }
};

Faster than 95.75%的

Mistakes：

要注意root value不是先插入的

***********上面思路的简化版本***********