145. Binary Tree Postorder Traversal

Description

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?

Idea

Use stack to iteratively traverse the binary tree.

Solution

/**
 * 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 {
public:
    // The order should be `left, right, root`
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> st;
        TreeNode *p = root;
        while (!st.empty() || p) {
            if (p) {
                st.push(p);
                // The order we push elements back to res is actually `root, right, left`
                res.push_back(p->val);
                p = p->right;
            } else {
                TreeNode *node = st.top();
                st.pop();
                p = node->left;
            }
        }
        
        // We need to reverse res
        reverse(res.begin(), res.end());
        
        return res;
    }
};