Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

  preorder是 root-left-right postorder是 left-right-root 那我们在preorder的基础上，改动push进stack的顺序，使得root-right-left 再翻转一下，也就是后pop出来的放进res vector的最前面，就实现left-right-root了   实现：

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        
        vector<int> res;
        if (!root) return res;
        
        stack<TreeNode*> st;
        st.push(root);
        while(!st.empty()){
            TreeNode* cur = st.top();
            st.pop();
            res.insert(res.begin(), cur->val);
            if (cur->left) st.push(cur->left);
            if (cur->right) st.push(cur->right);
        }
        
        return res;
        
    }
};