c语言输出二叉树先序遍历结果的第k个结点

可以使用递归的方法实现二叉树的先序遍历，并记录遍历到的结点个数。当遍历到第k个结点时，输出该结点的值即可。

以下是示例代码：

#include <stdio.h>
#include <stdlib.h>

// 二叉树结点定义
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

// 记录遍历到的结点个数
int count = 0;

// 先序遍历二叉树，返回第k个结点
struct TreeNode* preorder(struct TreeNode* root, int k) {
    if (root == NULL) {
        return NULL;
    }
    count++;
    if (count == k) {
        return root;
    }
    struct TreeNode* left = preorder(root->left, k);
    if (left != NULL) {
        return left;
    }
    struct TreeNode* right = preorder(root->right, k);
    if (right != NULL) {
        return right;
    }
    return NULL;
}

int main() {
    // 构造二叉树
    struct TreeNode* node1 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node2 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node3 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node4 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node5 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node1->val = 1;
    node2->val = 2;
    node3->val = 3;
    node4->val = 4;
    node5->val = 5;
    node1->left = node2;
    node1->right = node3;
    node2->left = node4;
    node2->right = NULL;
    node3->left = NULL;
    node3->right = node5;
    node4->left = NULL;
    node4->right = NULL;
    node5->left = NULL;
    node5->right = NULL;

    // 输出第3个结点的值，预期结果为4
    struct TreeNode* node = preorder(node1, 3);
    printf("%d\n", node->val);

    return 0;
}

输出结果为：

4