145. 二叉树的后序遍历

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

示例 1：

输入：root = [1,null,2,3]
输出：[3,2,1]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

提示：

• 树中节点的数目在范围 [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:

// 递归
/**
 * 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 postorder(struct TreeNode* root, int* res, int* returnSize) {
    if (root) {
        postorder(root->left, res, returnSize);
        postorder(root->right, res, returnSize);
        res[(*returnSize)++] = root->val;
    }
}
 
int* postorderTraversal(struct TreeNode* root, int* returnSize) {
    int* res = (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    postorder(root, res, returnSize);
    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* postorderTraversal(struct TreeNode* root, int* returnSize) {
    *returnSize = 0;
    int* res = (int*)malloc(sizeof(int) * 100);
    struct TreeNode* stack[100];
    int top = 0;
    struct TreeNode* p = root;
    struct TreeNode* flag = NULL; // 作为标记,等于刚刚访问过的元素,初始化为NULL
    while (p || top != 0) {
        if (p) {
            stack[top++] = p;
            p = p->left;
        } else {
            p = stack[top - 1]; // 获取栈顶元素
            if (p->right && p->right != flag) {
                // 栈顶元素有右子树且右子树还没有被访问过
                p = p->right;
                stack[top++] = p;
                p = p->left;
            } else {
                p = stack[--top];
                res[(*returnSize)++] = p->val;
                flag = p;
                p = NULL;
            }
        }
    }
    return res;
}