题解 | #牛牛的二叉树后序遍历#

/**
 * 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 back(TreeNode* root, vector<int>& res){
        if(root == nullptr) return;
        back(root->left, res);
        back(root->right, res);
        res.push_back(root->val);
        return;
    }
    
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        vector<int> res;
        back(root, res);
        return res;
    }
};