0145 - Binary Tree Postorder Traversal (Easy)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

// Time Complexity: O(N)
// Space Complexity: O(N)

// This is a standard post-order traversal problem, I'd suggest to learn in-order and pre-order as well.
// Here's a short tutorial if you're interested.
// https://wingkwong.github.io/leetcode-the-hard-way/tutorials/graph-theory/binary-tree
// then you may try the following problems
// 94. Binary Tree Inorder Traversal: https://leetcode.com/problems/binary-tree-inorder-traversal/
// 144. Binary Tree Postorder Traversal: https://leetcode.com/problems/binary-tree-preorder-traversal/

class Solution {
public:
    vector<int> ans;
    void postoder(TreeNode* node) {
        if (node == NULL) return;
        postoder(node->left);
        // traverse the left node
        postoder(node->right);
        // traverse the right node
        ans.push_back(node->val);
        // do something with node value here
    }

    vector<int> postorderTraversal(TreeNode* root) {
        postoder(root);
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
// Time complexity: O(n), where n - # of nodes in tree
// Space complexity: O(n)
class Solution {
    /**
     * Sample binary tree
     *
     *    1
     *   / \
     *  2   3
     *
     */
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        // Reference for last visited right node, for when parent is on top of the stack
        TreeNode last = null;
        while (root != null || !stack.isEmpty()) {
            // Keep pushing left nodes, all the way down onto stack
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                TreeNode node = stack.peek();
                // When Parent is on top stack, it checks with right node which has a refence in last variable
                // If both are same, it will not add repeated reference onto stack
                // Pops out stack top, i.e parent node, and level up higher for other nodes.
                if (node.right != null && node.right != last) {
                    root = node.right;
                } else {
                    // If any of the right node is empty, the block executes and add value from top of stack
                    result.add(node.val);
                    // Pops out stock top
                    last = stack.pop();
                }
            }
        }
        return result;
    }
}