LF.43.In-order Traversal Of Binary Tree(recur+iter)

Implement an iterative, in-order traversal of a given binary tree, return the list of keys of each node in the tree as it is in-order traversed.

Examples

5

/

3 8

/

1 4 11

In-order traversal is [1, 3, 4, 5, 8, 11]

Corner Cases

What if the given binary tree is null? Return an empty list in this case.

iterative:

public List<Integer> inOrder(TreeNode root) {
    // Write your solution here
    List<Integer> res = new ArrayList<>() ;
    Deque<TreeNode> stack = new LinkedList<>() ;
    //helper is the one-step ahead for the root
    TreeNode helper = root ;
    while(helper != null || !stack.isEmpty()){
        if (helper != null) {
            stack.offerFirst(helper);
            helper = helper.left ;
        } else{
            TreeNode curr = stack.pollFirst();
            res.add(curr.key) ;
            helper = curr.right ;
        }
    }
    return res ;
  }

recursive:

private List<Integer> preOrder_iter(TreeNode root){
        List<Integer> res = new ArrayList<>() ;
        if (root == null) {
            return res ;
        }
        helper(root, res);
        return res ;
    }
    private void helper(TreeNode root , List<Integer> res ){
            //base case
        if (root == null) {
            return ;
        }
        helper(root.left, res) ;
        res.add(root.key) ;
        helper(root.right, res);
    }