Leetcode Binary Tree Level Order Traversal-优快云博客

本文介绍了二叉树层级遍历的两种实现方法，通过使用队列来按层次顺序访问二叉树的所有节点，并返回每一层节点的值。第一种方法使用了两个队列交替进行层级遍历；第二种方法则是在一个队列中完成所有操作。

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Binary Tree Level Order Traversal

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

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

return its level order traversal as:

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

confused what"{1,#,2,3}"means?> read more on how binary tree is serialized on OJ.

利用两个队列会思路更加清晰点，增加了点额外空间。

class Solution {
public:
	vector<vector<int> > levelOrder(TreeNode *root) 
	{
		vector<vector<int> > v;
		if (!root) return v;

		queue<TreeNode *> qt1;
		qt1.push(root);
		queue<TreeNode *> qt2;
		
		vector<int> itmedia;
		itmedia.push_back(root->val);
		v.push_back(itmedia);
		itmedia.clear();

		while (!qt1.empty())
		{
			while (!qt1.empty())
			{
				TreeNode *tn = qt1.front();
				qt1.pop();
				if (tn->left)
				{
					qt2.push(tn->left);
					itmedia.push_back(tn->left->val);
				}
				if (tn->right)
				{
					qt2.push(tn->right);
					itmedia.push_back(tn->right->val);
				}
			}
			if (!qt2.empty())
			{
				qt1.swap(qt2);
				v.push_back(itmedia);
				itmedia.clear();
			}
		}
		return v;
	}
};

另外一种解法：

class Solution {
public:
	vector<vector<int> > levelOrder(TreeNode *root) 
	{
		vector<vector<int> > v;
		if (!root) return v;

		queue<TreeNode *> qt1;
		qt1.push(root);
		queue<TreeNode *> qt2;

		vector<int> itmedia;
		
		while (!qt1.empty())
		{
			while (!qt1.empty())
			{
				TreeNode *t = qt1.front();
				qt1.pop();
				itmedia.push_back(t->val);
				if (t->left) qt2.push(t->left);
				if (t->right) qt2.push(t->right);
			}
			v.push_back(itmedia);
			itmedia.clear();
			qt2.swap(qt1);
		}
		return v;
	}
};

//2014-2-16 update
	vector<vector<int> > levelOrder(TreeNode *root) 
	{
		vector<vector<int> > rs;
		if (!root) return rs;
		queue<TreeNode *> qt[2];
		qt[0].push(root);
		bool flag = false;
		while (!qt[flag].empty())
		{
			rs.push_back(vector<int>());
			while (!qt[flag].empty())
			{
				TreeNode *t = qt[flag].front();
				qt[flag].pop();
				rs.back().push_back(t->val);

				if (t->left) qt[!flag].push(t->left);
				if (t->right) qt[!flag].push(t->right);
			}
			flag = !flag;
		}
		return rs;
	}