# Three ways of Iterative PostOrder Traversing. Easy Explanation!

• Three types of Iterative Postorder Traversals.

1. Using 1 Stack. O(n) Time & O(n) Space
This is similar to Inorder using 1 Stack. The difference is we keep track of the previously printed node in `pre`. And we only print a node if its right child is `null` or equal to `pre`.
• Push all `left` nodes into the `stack` till it hits `NULL`.
• `root` = `s.peek()`
• if `root.right` = `null` or `pre` (Means we have traversed the right subtree already)
• We print `root` and pop it from `s`.
• Make `pre` = `root`
• `root` = `null` (So we dont go down to left child again)
• else
• `root` = `root.right` (Traverse the right subtree before printing `root`)
• Keep iterating till `both` the below conditions are met -
• Stack is empty `and`
• Root is NULL.
``````public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> out = new ArrayList<Integer>();
if(root==null)
return out;
TreeNode pre=null;
Stack<TreeNode> s = new Stack();
while(root!=null || !s.empty()){
if(root!=null){
s.push(root);
root = root.left;
}
else{
root = s.peek();
if(root.right==null || root.right==pre){
s.pop();
pre=root;
root = null;
}
else
root = root.right;
}
}
return out;
}
``````
1. Using 2 Stacks. O(n) Time & O(n) Space
We use two stacks. Stack `s` is used to find and traverse the child nodes, and `path` stack keeps track of the path from the `root` to the current node. (This is usefull in certain problems like Binary Tree Paths and Path Sum ).
• Initially we push the `root` into `s`.
• Keep iterating with below logic till `s` is `empty`.
• `root` = `s.peek()`
• If the top elements of both the stacks are not the same :
• Print `root` and push it into `path`.
• Push `root`'s children into `s` in reverse order. (Remember it's a stack!)
• When top elements of both stacks are equal. (Which means we hit a deadend, and need to turn back)
• `Pop` from `both` stacks.
``````public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> out = new ArrayList<Integer>();
if(root == null)
return out;
Stack<TreeNode> s = new Stack(), path = new Stack();
s.push(root);
while(!s.empty()){
root = s.peek();
if(!path.empty() && path.peek()==root){
s.pop();
path.pop();
}
else{
path.push(root);
if(root.right != null)
s.push(root.right);
if(root.left != null)
s.push(root.left);
}
}
return out;
}
``````
1. Using No Stacks (Morris Traversal). O(n) Time & O(1) Space
Instead of using stacks to remember our way back up the tree, we are going to modify the tree to create upwards links. The idea is based on Threaded Binary Tree.
Similar to Inorder Morris Traversal.

We reverse each diagonal shown in the picture (d1-d4), print it and re-reverse it.

• Create a `dummy` node and make `dummy.left` = `root`.
• `root` = `dummy`
• Iterate till `root` is null.
• If `root` has a left child.
• Find the `inorder predecessor` => `pre`. (Inorder predecessor of root is the right most child of its left child)
• `pre.right` = `root` (Make it point to root).
• `root` = `root.left`.
• If its already pointing to root (which means we have traversed it already and are on our way up.)
• Reverse from `root.left` to `pre`.
• Traverse from `pre` to `root.left` and print the nodes.
• Re-reverse it back to normal.
• `pre.right` = `null`.
• `root` = `root.right`.
• If left child is `null`
• `root` = `root.right`. (We are climbing up our link.)
``````public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> out = new ArrayList<Integer>();
if(root == null)
return out;
TreeNode dummy = new TreeNode(-1), pre = null;
dummy.left = root; root = dummy;
while(root != null){
if(root.left != null){
pre = root.left;
while(pre.right != null && pre.right != root)
pre=pre.right;
if(pre.right == null){
pre.right = root;
root = root.left;
}
else{
TreeNode node = pre;
reverse(root.left,pre);
while(node != root.left){
node = node.right;
}
out.add(node.val);          // Print again since we are stopping at node=root.left
reverse(pre,root.left);
pre.right = null;
root = root.right;
}
}
else{
root = root.right;
}
}
return out;
}

public void reverse(TreeNode from, TreeNode to){
if(from == to)
return;
TreeNode prev = from, node = from.right;
while(prev != to){
TreeNode next = node.right;
node.right = prev;
prev = node;
node = next;
}
}
``````

Also checkout Inorder & PreOrder :))

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