Reverse Level Order Traversal

Problem Statement

Pattern:

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Once you know Level Order Traversal ,

Solution

public ArrayList<Integer> reverseLevelOrder(Node root) {
	if(root == null) return new ArrayList<Integer>();
	// result list
	ArrayList<Integer> res = new ArrayList<>();
	
	// queue to iterate over BT
	Queue<Node> q = new LinkedList<Node>();
	q.add(root);
	while (!q.isEmpty()) {
		Node node = q.poll();
		res.add(node.val);
		
		if(node.right != null) q.add(node.right);
		if(node.left != null) q.add(node.left);
	}
	Collections.reverse(res);
	return res;
}

ArrayList Solution

Binary Tree Level Order Traversal II - LeetCode

public List<List<Integer>> levelOrderBottom(TreeNode root) {
	
	List<List<Integer>> res = new ArrayList<>();
	if(root == null) return res;
 
	Queue<TreeNode> q = new LinkedList<TreeNode>();
	q.add(root);
	while (!q.isEmpty()) {
		int size = q.size();
		ArrayList<Integer> level = new ArrayList<>();
		while(size-- > 0) {
			TreeNode node = q.poll();
			level.add(node.val);
			if(node.left != null) q.add(node.left); 
			if(node.right != null) q.add(node.right); 
		}
		res.add(level);
	}
	Collections.reverse(res);
	return res;
}

Notes

• This Question is a slight modification of Level Order Traversal using an Iterative Queue: ![[Binary Tree#Level Order Traversals#Iterative Queue O n]]
• In this one
  • Add the right node to the queue first, then the left node
  • In the end we get an ArrayList with
    • levels in fwds
    • nodes in a level are in reverse
  • now if we reverse the res ArrayList we get
    • all the levels in reverse
    • nodes in level are fwds