144. 二叉树的前序遍历

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

示例 4：

输入：root = [1,2]
输出：[1,2]

示例 5：

输入：root = [1,null,2]
输出：[1,2]

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

进阶：递归算法很简单，你可以通过迭代算法完成吗？

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
 
void preorder(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL)
        return;
    res[(*returnSize)++] = root->val;
    preorder(root->left, res, returnSize);
    preorder(root->right, res, returnSize);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int* res = (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    preorder(root, res, returnSize);
    return res;
}

// 精简后的代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
 
void preorder(struct TreeNode* root, int* returnSize, int* res) {
    if (root) {
        res[(*returnSize)++] = root->val;
        preorder(root->left, returnSize, res);
        preorder(root->right, returnSize, res);
    }
}
 
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    *returnSize = 0;
    int* res = (int*)malloc(sizeof(int) * 1000);
    preorder(root, returnSize, res);
    return res;
}

// 非递归,即循环/迭代
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    struct TreeNode* p = root;
    struct TreeNode* q = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* stack[100];
    int top = 0;
    *returnSize = 0;
    int* res = (int*)malloc(sizeof(int) * 100);
    while (p || top != 0) {
        if (p) {
            stack[top++] = p;
            res[(*returnSize)++] = p->val; // 在此处访问根结点
            p = p->left;
        } else {
            q = stack[--top];
            p = q->right;
        }
    }
    return res;
}