题解 | #二叉树的后序遍历#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
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) {
        // write code here
        vector<int> result;

        if (!root) return result;

        vector<int> left_sub_tree = postorderTraversal(root->left);
        result.insert(result.end(), left_sub_tree.begin(), left_sub_tree.end());
        vector<int> right_sub_tree = postorderTraversal(root->right);
        result.insert(result.end(), right_sub_tree.begin(), right_sub_tree.end());

        result.push_back(root->val);


        return result;
    }
};