Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Solution: BFS
进行一次广度优先搜索，维护一个队列，只是对于每个结点它的邻接点只有可能是左孩子和右孩子。
算法的复杂度是就结点的数量O(n)，空间复杂度是一层的结点数，也是O(n)。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList> levelOrder(TreeNode root) {
        ArrayList> result = new ArrayList>();
        
        if(root == null)
            return result;
        
        //use queue to store nodes
        LinkedList queue = new LinkedList();
        queue.add(root);
        
        int curLevCount = 1;   //nodes number in current level
        int nextLevCount = 0;   //nodes number in next level
        
        ArrayList levelRes = new ArrayList();
        
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            curLevCount--;
            levelRes.add(cur.val);
            
            if(cur.left != null) {   //store current's left subnode into queue
                queue.add(cur.left);
                nextLevCount++;
            }
            
            if(cur.right != null) {   //store current's right subnode into queue
                queue.add(cur.right);
                nextLevCount++;
            }
            
            if(curLevCount == 0) {
                curLevCount = nextLevCount;
                nextLevCount = 0;
                result.add(levelRes);
                levelRes = new ArrayList();
            }
        }
        return result;
    }
}