C 解法, 执行用时: 2ms, 内存消耗: 396KB, 提交时间: 2022-05-29

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @return int整型一维数组
 * @return int* returnSize 返回数组行数
 */
void postOrder(struct TreeNode* node, int* ret, int* returnSize) {
    if(!node) return;
    postOrder(node->left, ret, returnSize);
    postOrder(node->right, ret, returnSize);
    ret[(*returnSize)++] = node->val;
}
 
int* postorderTraversal(struct TreeNode* root, int* returnSize){
    int* ret= (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    postOrder(root, ret, returnSize);
    return ret;
}

C++ 解法, 执行用时: 2ms, 内存消耗: 468KB, 提交时间: 2022-07-24

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector
     */
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        vector<int> res;
        TreeNode *t = root, *r=nullptr;

        while(!s.empty() || t) {
            while(t) {
                s.push(t);
                t = t->left;
            }
            t = s.top();
            if (t->right && t->right!=r) {
                t = t->right;
                continue;
            }
            res.push_back(t->val);
            r = t;
            s.pop();
            t = nullptr;
        }
        return res;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 472KB, 提交时间: 2022-08-06

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector
     */
    void postorder(TreeNode* root, vector<int>& res) {
        if (root != nullptr) {
            postorder(root->left, res);
            postorder(root->right, res);
            res.push_back(root->val);
        }
    }
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        vector<int> res;
        postorder(root, res);
        return res;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 540KB, 提交时间: 2022-04-17

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector
     */
    vector<int> res;
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        if (root == nullptr) return {};
        dfs(root);
        return res;
    }
    
    void dfs(TreeNode* node) {
        if (node == nullptr) return;
        dfs(node->left);
        dfs(node->right);
        res.push_back(node->val);
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 580KB, 提交时间: 2022-03-13

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector
     */
    vector<int> backvec;
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        if(root != nullptr){
            backOrder(root);
        }
        return backvec;
    }
    void backOrder(TreeNode* node){
        if(node == nullptr){
            return;
        }
        backOrder(node->left);
        backOrder(node->right);
        backvec.push_back(node->val);
    }
};