LeetCode //C - 230. Kth Smallest Element in a BST

230. Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k , return the k t h k^{th} kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

Input: root = 3,1,4,null,2, k = 1
Output: 1

Example 2:

Input: root = 5,3,6,2,4,null,null,1, k = 3
Output: 3

Constraints:

• The number of nodes in the tree is n.
• 1 < = k < = n < = 1 0 4 1 <= k <= n <= 10^4 1<=k<=n<=104
• 0 < = N o d e . v a l < = 1 0 4 0 <= Node.val <= 10^4 0<=Node.val<=104

From: LeetCode

Link: 230. Kth Smallest Element in a BST

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Solution:

Ideas：

Here's how the algorithm will work:

1. Perform an in-order traversal of the BST.
2. As you traverse the nodes, keep a count of the nodes visited.
3. When the count becomes equal to k, that node's value is the kth smallest. current node and the previous node.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
typedef struct {
    int count;
    int value;
} Result;

void inOrderTraversal(struct TreeNode* root, int k, Result* res) {
    if (root == NULL || res->count == k) return;

    inOrderTraversal(root->left, k, res);
    if (++res->count == k) {
        res->value = root->val;
    }
    inOrderTraversal(root->right, k, res);
}

int kthSmallest(struct TreeNode* root, int k) {
    Result res = {0, 0};
    inOrderTraversal(root, k, &res);
    return res.value;
}