Day20. KthSmallestElementInBST.cpp

/*
Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1
Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

   Hide Hint #1  
Try to utilize the property of a BST.
   Hide Hint #2  
Try in-order traversal. (Credits to @chan13)
   Hide Hint #3  
What if you could modify the BST node's structure?
   Hide Hint #4  
The optimal runtime complexity is O(height of BST).

*/

/**
 * 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:
    
    TreeNode* help(TreeNode* root,int k,int& count){
        if(root==NULL) return NULL;
        
        TreeNode* l=help(root->left,k,count);
        
        if(l!=NULL) return l;
        
        count++;
        
        if(count==k) return root;
        
        return help(root->right,k,count);
        
        
    }
    int kthSmallest(TreeNode* root, int k) {
        int count=0;
        TreeNode* res=help(root,k,count);
        if(res==NULL) return -1;
        
        else return res->val;
    }
};