[leetcode] 272. Closest Binary Search Tree Value II 解题报告

题目链接: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:

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.

思路: 利用二叉搜索树的中序遍历, 如果结果集合中元素还不到k个, 就把当前元素加到集合中去, 如果集合中的元素个数多于k了, 那么有二种情况:

1. target的值比集合中最小的值小, 因为中序遍历是有序的, 最小元素就是第0个元素

2. target的值大于结果集合中的最小值, 小于最大值

3. target的值大于集合中的最大值

当集合中的元素个数大于k的时候, 我们需要进行元素替换. 

如果是第一种情况, 那么将无法进行替换, 因为第0个元素就是最靠近的值, 并且在集合中元素是有序的, 因此此时集合中就是最靠近的k个元素.

如果是第二三种情况, 那么我们比较当前值和集合中最小值, 如果当前结点值比那个值更靠近target, 那么我们就用当前元素替换最小的值

直到我们无法再找到能够替换的元素, 就可以返回结果了.

时间复杂度小于O(N)

代码如下:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> closestKValues(TreeNode* root, double target, int k) {
        if(!root) return result;
        closestKValues(root->left, target, k);
        if(result.size() < k) result.push_back(root->val);
        else
        {
            if(fabs(root->val-target) < fabs(result[index]-target))
            {
                result[index] = root->val;
                index++;
                if(k == index) index = 0;
            }
            else return result;
        }
        closestKValues(root->right, target, k);
        return result;
    }
private:
    vector<int> result;
    int index =0;
};

参考: https://leetcode.com/discuss/83431/c-simple-inorder-solution-with-o-n-runtime-and-o-k-memory