LeetCode 107:Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

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

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

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

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//解题思路：分层遍历二叉树，用队列模拟。
class Solution {
public:
	vector<vector<int> > levelOrderBottom(TreeNode *root) {
		vector< vector<int> > result;
		if (root == NULL)
			return result;

		queue<TreeNode*> q; 
		q.push(root);
		int count = 1; //count表示当前层的节点个数
		int level = 0;
		vector<int> tmp(0); //tmp用来临时保存当前层的节点
		while (!q.empty())
		{
			tmp.clear();
			level = 0;
			for (int i = 0; i < count; ++i)
			{
				root = q.front();
				q.pop();
				tmp.push_back(root->val);
				if (root->left != NULL)  
				{
					q.push(root->left);  //队列q用来保存下一层的节点
					++level;
				}
				if (root->right != NULL)
				{
					q.push(root->right);
					++level;
				}
			}
			count = level;
			result.push_back(tmp);
		}

		reverse(result.begin(), result.end());
		return result;
	}
};