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 <algorithm>
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) {
        vector<int> res;
        if (!root) return res;
        stack<TreeNode*> sta;
        sta.push(root);
        while (!sta.empty()) {
            TreeNode* tmp = sta.top();
            sta.pop();
            res.push_back(tmp->val); 
            if (tmp->left) sta.push(tmp->left);
            if (tmp->right) sta.push(tmp->right);
           
        }
        reverse(res.begin(),res.end());
        return res;
    }
};