[LeetCode] Kth Smallest Element in a BST

Problem:

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
https://leetcode.com/problems/kth-smallest-element-in-a-bst/

Thinking:

The kth smallest node is the left child's left child...(all the way to leaf node) of the right child of the (k - 1) th node in a binary search tree.

参考：https://leetcode.com/discuss/43299/o-k-space-o-n-time-10-short-lines-3-solutions

C++ Code：(脑洞：刚开始写C++，代码风格好像怪怪的)

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> nodes;
        if(root == NULL) return -1;
        while(true){
            while(root!=NULL){
                nodes.push(root);
                root=root->left;
            }
            TreeNode* node=nodes.top();
            nodes.pop();
            if(k==1) return node->val;
            else {
                root=node->right;
                k--;
            }
        }
    }
};

Java Code (Using Divide and Conquer):

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int leftCount = countNodes(root.left) + 1;
        if (leftCount == k) {
            return root.val;
        } else if (leftCount > k) {
            return kthSmallest(root.left, k);
        } else {
            return kthSmallest(root.right, k - leftCount);
        }
    }
    
    private int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}