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).

Have you met this question in a real interview?

Yes

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]
]

Challenge

Challenge 1: Using only 1 queue to implement it.

Challenge 2: Use DFS algorithm to do it.

Tips: C# Queue 操作 http://www.cnblogs.com/tianzhiliang/archive/2010/09/21/1832664.html

定理： Queue 和 BFS搭配使用; Stack 和 DFS搭配使用.

IList<IList<int>> result = new List<IList<int>>();

把树的node扔进queue, 然后循环输出， queue 先进先出

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<IList<int>> LevelOrder(TreeNode root) {                 
        IList<IList<int>> result = new List<IList<int>>();

        if(root == null){
            return result;
        }
        //1: 创建一个队列把启示节点放入
        Queue<TreeNode> queue = new Queue<TreeNode>();
        queue.Enqueue(root);

        //2. while 队列不空，处理队列的节点并拓展出新的节点
        while (queue.Count!=0){
            IList<int> level = new List<int>();

            int count = queue.Count;
            for (int i = 0; i < count;i++){
                TreeNode head = queue.Dequeue();                    
                level.Add(head.val); 

                if (head.left != null){
                    queue.Enqueue(head.left);
                }
                if (head.right != null){
                    queue.Enqueue(head.right);
                }
            }

            result.Add(level);
        }
        return result;
    }
}