145. Binary Tree Postorder Traversal (Hard)

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

For example: Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

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

postorder: left-right-middle

Solution 1: recursive

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    void postOrder(TreeNode *root, vector<int> &res) {
        if (root) {
            postOrder(root->left, res);
            postOrder(root->right, res);
            res.push_back(root->val);
        }
    }
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        postOrder(root, res);
        return res;
    }
};

Solution 2: iterative

vector<int> postorderTraversal(TreeNode* root) {
    vector<int> res;
    if (!root) return res;

    stack<TreeNode*> s;
    s.push(root);
    TreeNode *head = root;
    while (!s.empty()) {
        TreeNode* top = s.top(); 
        // push node if it has no children, or its children has been visited
        if ((!top->left && !top->right) || top->left == head || top->right == head) {
            res.push_back(top->val);
            s.pop();
            head = top;
        } else {
            if (top->right) s.push(top->right);
            if (top->left) s.push(top->left); // last in fist out
            // test left node first
        }
    }
    return res;
}