Leetcode Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:

Given binary tree

{3,9,20,#,#,15,7}

,

3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

我使用了两个deque容器做，其实可以使用stack。

代码还是非常清晰的，一块一块的分好，思路很容易跟的。

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

		deque<TreeNode *> qt1;
		deque<TreeNode *> qt2;
		qt1.push_back(root);

		vector<int> itmedia;
		itmedia.push_back(root->val);
		v.push_back(itmedia);
		itmedia.clear();

		while (!qt1.empty())
		{
			while (!qt1.empty())
			{
				TreeNode *t = qt1.back();
				qt1.pop_back();
				if (t->right)
				{
					qt2.push_back(t->right);
					itmedia.push_back(t->right->val);
				}
				if (t->left)
				{
					qt2.push_back(t->left);
					itmedia.push_back(t->left->val);
				}
			}
			if (!itmedia.empty()) v.push_back(itmedia);
			itmedia.clear();
			while (!qt2.empty())
			{
				TreeNode *t = qt2.back();
				qt2.pop_back();
				if (t->left)
				{
					qt1.push_back(t->left);
					itmedia.push_back(t->left->val);
				}
				if (t->right)
				{
					qt1.push_back(t->right);
					itmedia.push_back(t->right->val);
				}
			}
			if (!itmedia.empty()) v.push_back(itmedia);
			itmedia.clear();
		}
		return v;
	}
};