Get the list of list of keys in a given binary tree layer by layer. Each layer is represented by a list of keys and the keys are traversed from left to right.

Examples

5

/    \

3        8

/   \        \

1     4        11

the result is [ [5], [3, 8], [1, 4, 11] ]

Corner Cases

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

How is the binary tree represented?

We use the level order traversal sequence with a special symbol "#" denoting the null node.

For Example:

The sequence [1, 2, 3, #, #, 4] represents the following binary tree:

1

/   \

2     3

/

4

/**
* public class TreeNode {
* public int key;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int key) {
* this.key = key;
* }
* }
*/
public class Solution {
public List<List<Integer>> layerByLayer(TreeNode root) {
// Write your solution here
// use BFS to solve this, thus, we use queue to maintain the nodes that we have seen but haven't deal with,
// and for each layer, we maintain a List to store all keys in this layer
// the tree can be null, then we return a List contains nothing
// the number of the elements in a single layer is less than Integer.MAX_VALUE
List<List<Integer>> res = new ArrayList<>();
if(root==null){
return res;
}
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
while(q.size()>0){
int size = q.size();
List<Integer> layer = new ArrayList<>();
for(int i=0; i<size; i++){
TreeNode cur = q.poll();
if(cur!=null){
layer.add(cur.key);
}
if(cur.left!=null){
q.offer(cur.left);
}
if(cur.right!=null){
q.offer(cur.right);
}
}
res.add(layer);
}
return res;
}
}

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