Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

 Total Accepted: 4998 Total
Submissions: 16087My Submissions

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:

Given binary tree 

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7]
[9,20],
[3],
]

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

BFS

public static ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root){
if(root == null) return new ArrayList<>();
ArrayList<ArrayList<Integer>> totalList = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while(!queue.isEmpty()){
Queue<TreeNode> subQ = new LinkedList<>();
ArrayList<Integer> subList = new ArrayList<>();
while(!queue.isEmpty()){
TreeNode node = queue.poll();
subList.add(node.val);
if(node.left!=null) subQ.add(node.left);
if(node.right!=null) subQ.add(node.right);
}
totalList.add(subList);
queue = subQ;
}
Collections.reverse(totalList);
return totalList;
}