题解 | #二叉树的后序遍历#C++暴力递归解法

/**
 * 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 answer(TreeNode*root,vector <int> &ans)
     {
        if(root)
        {
            answer(root->left,ans);
            answer(root->right,ans);
            ans.push_back(root->val);
        }
     }
    vector<int> postorderTraversal(TreeNode* root) 
    {
        vector <int> ans;
        answer(root,ans);
        return ans;
    }
};