leetcode 102 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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

给定一个二叉树，按层次输出其节点，而且输出的结果需体现其节点的层次信息。二叉树的问题基本都可以使用深度优先搜索或广度优先搜索来解决，本题就是典型的广度优先搜索问题。核心是遍历到新的层次时，需要压入一个新的数组，以后同一层级的节点都按左右顺序压入，递归版本代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
		vector<vector<int>> ret;
		bfsNodes(ret, 0, root);
		return ret;
	}

	void bfsNodes(vector<vector<int>> &ret, int level,  TreeNode* root)
	{
		if (NULL == root)
		{
			return;
		}
		if (ret.size() == level)//Attension: push new vector when traverse the most left node
		{
			ret.push_back({});
		}
		ret[level].push_back(root->val);//ret aready has element at level
		if (root->left)
		{
			bfsNodes(ret, level + 1, root->left);
		}
		if (root->right)
		{
			bfsNodes(ret, level + 1, root->right);
		}
	}
};

二叉树的遍历还有非递归的方式，这种方式是之前缺少练习的。这种情况下需要借助一个队列，遍历每层的节点时，在该层遍历该层节点数，每遍历到一个节点时，将该节点弹出，记录下该节点的值，同时压进左右子节点，这样遍历下一层的节点时下层的节点数便是队列的长度，如此直到队列为空。非递归版本代码如下：

class Solution {
public:
	vector<vector<int>> levelOrder(TreeNode* root) {
		vector<vector<int>> ret;
		if (!root)
		{
			return ret;
		}
		queue<TreeNode*> q_node;
		q_node.push(root);
		while (!q_node.empty())
		{
			vector<int> v_level;
			int curr_size = q_node.size();	//record the number of current level
			for (int i = 0; i < curr_size; i++)
			{
				TreeNode* tmp_node = q_node.front();
				q_node.pop();
				if (tmp_node)
				{
					v_level.push_back(tmp_node->val);
					if (tmp_node->left)
					{
						q_node.push(tmp_node->left);
					}
					if (tmp_node->right)
					{
						q_node.push(tmp_node->right);
					}
				}
			}
			ret.push_back(v_level);
		}
		return ret;
	}
};