Binary Tree Postorder Traversal.java

/*
    Given a binary tree, return the postorder traversal of its nodes' values.
    
    Link: https://leetcode.com/problems/binary-tree-postorder-traversal/

    Example: For example:
		Given binary tree {1,#,2,3},
		   1
		    \
		     2
		    /
		   3
		return [3,2,1].

    Solution: None
    
    Source: None
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> lst = new ArrayList<Integer>();

        if (root == null)
            return lst; 

        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);

        TreeNode prev = null;
        while (! stack.empty()) {
            TreeNode curr = stack.peek();

            // go down the tree.
            //check if current node is leaf, if so, process it and pop stack,
            //otherwise, keep going down
            if (prev == null || prev.left == curr || prev.right == curr) {
                //prev == null is the situation for the root node
                if (curr.left != null) {
                    stack.push(curr.left);
                } else if (curr.right != null) {
                    stack.push(curr.right);
                } else {
                    stack.pop();
                    lst.add(curr.val);
                }

            //go up the tree from left node    
            //need to check if there is a right child
            //if yes, push it to stack
            //otherwise, process parent and pop stack
            } else if (curr.left == prev) {
                if (curr.right != null) {
                    stack.push(curr.right);
                } else {
                    stack.pop();
                    lst.add(curr.val);
                }

            //go up the tree from right node 
            //after coming back from right node, process parent node and pop stack. 
            } else if (curr.right == prev) {
                stack.pop();
                lst.add(curr.val);
            }

            prev = curr;
        }

        return lst; 
    }
}