Binary Tree Non-recursive Traversal

    public static void BSTPreorderTraverse(Node node) {
        if (node == null) {
            return;
        }
        Stack<Node> s = new Stack<Node>();
        s.push(node);
        while (!s.empty()) {
            node = s.pop();
            System.out.println(node.toString());
            if (node.rightChild != null) {s.push(node.rightChild);}
            if (node.leftChild != null) {s.push(node.leftChild);}
        }
    }

 1 public static void BSTInorderTraverse(Node node) {
 2     Stack<Node> s = new Stack<Node>();
 3     while (!s.empty() || node != null) {
 4         if (node != null) {
 5             s.push(node);
 6             node = node.leftChild;
 7         } else {
 8             node = s.pop();
 9             System.out.println(node.toString());
10             node = node.rightChild;
11         }
12     }
13 }

 1 public List<Integer> postorderTraversal(TreeNode root) {
 2         List<Integer> res = new ArrayList<Integer>();
 3 
 4         if (root == null) {
 5             return res;
 6         }
 7 
 8         Stack<TreeNode> stack = new Stack<TreeNode>();
 9         stack.push(root);
10 
11         while (!stack.isEmpty()) {
12             TreeNode temp = stack.peek();
13             if (temp.left == null && temp.right == null) {
14                 TreeNode pop = stack.pop();
15                 res.add(pop.val);
16             } else {
17                 if (temp.right != null) {
18                     stack.push(temp.right);
19                     temp.right = null; // if we don‘t want to change the tree structure, we can use a set to check whether the node‘s children have been added to the stack or not.
20                 }
21 
22                 if (temp.left != null) {
23                     stack.push(temp.left);
24                     temp.left = null;
25                 }
26             }
27         }
28 
29         return res;
30     }