题解 | #二叉树的后序遍历#

后序遍历：左-右-根。果然还是递归写起来最简单🙂

/**
 * 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) {
        // write code here
        vector<int> res;
        if (!root) return res;
        vector<int> l = postorderTraversal(root->left),
                    r = postorderTraversal(root->right);
        for (int i = 0; i < l.size(); i++) {
            res.push_back(l[i]);
        }
        for (int i = 0; i < r.size(); i++) {
            res.push_back(r[i]);
        }
        res.push_back(root->val);
        return res;
    }
};