Day 19:LeetCode 30 Day may challenges

Kth_smallest_element_in_BST.cpp

/**
 * 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) {}
 * };
 */
///// METHOD-1 /////
// best in space complexity
class Solution {
public:
    //       5
    //      /  \
    //      4   7
    //    
    //
    int l = 0;
    int ans;
    void Inorder(TreeNode* root,int& k){
        if(root == NULL)
            return;
        Inorder(root->left,k);
        //cout<<root->val;
        l+=1;
        if(l == k)
            ans = root->val;
        Inorder(root->right,k);    
    }
    int kthSmallest(TreeNode* root, int k) {
        //my first idea is inorder traversal
        Inorder(root,k);
        return ans;
    }
};
////// METHOD-2 //////
class Solution {
public:
    //       5
    //      /  \
    //      4   7
    //    
    //
    vector<int> v;
    void Inorder(TreeNode* root){
        if(root == NULL)
            return;
        Inorder(root->left);
        //cout<<root->val;
        v.push_back(root->val);
        Inorder(root->right);    
    }
    int kthSmallest(TreeNode* root, int k) {
        //my first idea is inorder traversal
        Inorder(root);
        return v[k-1];
    }
};

Problem_statement19.txt

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).