Get Keys In Binary Tree Layer By Layer

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;
  }
}

来源：https://www.cnblogs.com/zg1005/p/12131316.html