/*
    Given root of BST & int k, return kth smallest value (1-indexed) of all values in tree
    Ex. root = [3,1,4,null,2] k = 1 -> 1, root = [5,3,6,2,4,null,null,1] k = 3 -> 3

    Inorder traversal, each visit decrement k, when k = 0 return, works because inorder

    Time: O(n)
    Space: O(n)
*/

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        int result = 0;
        inorder(root, k, result);
        return result;
    }
private:
    void inorder(TreeNode* root, int& k, int& result) {
        if (root == NULL) {
            return;
        }
        inorder(root->left, k, result);
        k--;
        if (k == 0) {
            result = root->val;
            return;
        }
        inorder(root->right, k, result);
    }
};