Three ways of Iterative Preorder Traversing. Easy Explanation


  • 4

    Three types of Iterative Preorder Traversals in java.

    1. Using 1 Stack. O(n) Time & O(n) Space
      • Print and push all left nodes into the stack till it hits NULL.
      • Then Pop the top element from the stack, and make the root point to its right.
      • Keep iterating till both the below conditions are met -
        • Stack is empty and
        • Root is NULL.
    public List<Integer> preorderTraversal(TreeNode root) {
    	List<Integer> out = new ArrayList<Integer>();
    	if(root==null)
    		return out;
    	Stack<TreeNode> s = new Stack();      
    	while(root!=null || !s.empty()){
    		if(root!=null){
    			out.add(root.val);
    			s.push(root);
    			root = root.left;
    		}
    		else{
    			root = s.pop();				
    			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{
                out.add(root.val);
                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.
      • Iterate till root is null.
        • If root has a left child.
          • Find the inorder predecessor. (Inorder predecessor of root is the right most child of its left child)
            • 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.)
            • Make the inorder predecessor point to null (Reverting our structural changes)
            • root = root.right.
        • If left child is null
          • root = root.right. (We are climbing up our link.)
    public List<Integer> preorderTraversal(TreeNode root) {
    	List<Integer> out = new ArrayList<Integer>();
    	if(root == null)
    		return out;
    	TreeNode pre = null;
    	while(root!=null){
    		if(root.left !=null){
    			pre = root.left;
    				while(pre.right!=null && pre.right!=root)
    				pre=pre.right;
    			if(pre.right==null){
    				out.add(root.val);
    				pre.right=root;
    				root=root.left;
    			}
    			else{
    				pre.right=null;
    				root=root.right;
    			}                   
    		}
    		else{
    			out.add(root.val);
    			root=root.right;
    		}
    	}
    	return out;
    }
    

    Also checkout Inorder & PostOrder :))


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