145. Binary Tree 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?

to get post-order, you will need 2 stacks to get the result. in the question. the result array is the second stack.

/**
 * 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> res = new ArrayList<>();
        if(root == null) return res;
        Stack<TreeNode> stack1 = new Stack<>();
        stack1.push(root);
        while(!stack1.isEmpty()){
            TreeNode node = stack1.pop();

            res.add(node.val);
            if(node.left != null) stack1.push(node.left);
            if(node.right != null) stack1.push(node.right);

        }

        Collections.reverse(res);

        return res;

    }
}