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<int> closestKValues(TreeNode* root, double target, int k) {stack<TreeNode*> suc, pre;vector<int> 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<TreeNode*>& 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<TreeNode*>& 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<TreeNode*>& 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<TreeNode*>& 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;}};