Leetcode 272. Closest Binary Search Tree Value II (Hard) (cpp)

Leetcode 272. Closest Binary Search Tree Value II (Hard) (cpp)

Tag: Tree, Stack

Difficulty: Hard

/*

272. Closest Binary Search Tree Value II (Hard)

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.

*/
/**
* 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 closestKValues(TreeNode* root, double target, int k) {
		stack suc, pre;
		vector res;
		iniPreSta(root, target, pre);
		iniSucSta(root, target, suc);
		if (!suc.empty() && !pre.empty() && suc.top()->val == pre.top()->val) {
			getNextPredecessor(pre);
		}
		while (k-- > 0) {
			if (suc.empty()) {
				res.push_back(getNextPredecessor(pre));
			}
			else if (pre.empty()) {
				res.push_back(getNextSuccessor(suc));
			}
			else {
				if (abs((double)pre.top()->val - target) < abs((double)suc.top()->val - target)) {
					res.push_back(getNextPredecessor(pre));
				}
				else {
					res.push_back(getNextSuccessor(suc));
				}
			}
		}
		return res;
	}
private:
	void iniPreSta(TreeNode* root, double target, stack& pre) {
		while (root != NULL) {
			if (root->val < target) {
				pre.push(root);
				root = root->right;
			}
			else if (root->val > target) {
				root = root->left;
			}
			else {
				pre.push(root);
				break;
			}
		}
	}
	void iniSucSta(TreeNode* root, double target, stack& suc) {
		while (root != NULL) {
			if (root->val > target) {
				suc.push(root);
				root = root->left;
			}
			else if (root->val < target) {
				root = root->right;
			}
			else {
				suc.push(root);
				break;
			}
		}
	}
	int getNextSuccessor(stack& suc) {
		TreeNode* cur = suc.top();
		suc.pop();
		int res = cur->val;
		cur = cur->right;
		while (cur != NULL) {
			suc.push(cur);
			cur = cur->left;
		}
		return res;
	}
	int getNextPredecessor(stack& pre) {
		TreeNode* cur = pre.top();
		pre.pop();
		int res = cur->val;
		cur = cur->left;
		while (cur != NULL) {
			pre.push(cur);
			cur = cur->right;
		}
		return res;
	}
};