题目

https://oj.leetcode.com/submissions/detail/10496925/

分析

树的层次遍历， 唯一的难点在于如何判断两个结点是否在同一层上。

可以除了nodeQue之外，再增加一个depQue, 来记录每个结点的深度。

代码：

class Solution
{
public:
	vector<vector<int>> levelOrder(TreeNode *root)
	{
		queue<TreeNode*> nodeQue;
		queue<int> depQue;
		TreeNode *node;
		int dep;
		int depBefore;
		vector<vector<int>> res;
		vector<int> solution;

		if (!root)
			return res;

		nodeQue.push(root);
		depQue.push(1);
		depBefore = 1;
		while (!nodeQue.empty())
		{
			node = nodeQue.front();
			nodeQue.pop();
			dep = depQue.front();
			depQue.pop();

			if (dep == depBefore)
				solution.push_back(node->val);
			else
			{
				res.push_back(solution);
				depBefore = dep;
				solution.clear();
				solution.push_back(node->val);
			}

			if (node->left)
			{
				nodeQue.push(node->left);
				depQue.push(dep + 1);
			}
			if (node->right)
			{
				nodeQue.push(node->right);
				depQue.push(dep + 1);
			}
		}
		res.push_back(solution);
		return res;
	}
};

本文章迁移自http://blog.csdn.net/timberwolf_2012/article/details/38927101