Postorder Traversal

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?

Solution

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack, result = [], []
        stack.append(root)
        while stack:
            node = stack.pop()
            result.append(node.val)
            if node.left is not None: stack.append(node.left)
            if node.right is not None: stack.append(node.right)
        result.reverse()
        return result

/**
 * 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) {
        LinkedList<Integer> result = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();

        if(root == null)
            return result;

        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            result.addFirst(cur.val);

            if(cur.left != null)
                stack.push(cur.left);
            if(cur.right != null)
                stack.push(cur.right);
        }

        return result;

    }
}