Binary Tree Level Order Traversal II @leetcode

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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.

方法很囧，就是把1里面的产生的vector res给reverse一下就行。几乎没什么改变。。不知道有木有不用reverse的办法。

	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	vector<vector<int> > levelOrderBottom(TreeNode *root) {
	// Start typing your C/C++ solution below
	// DO NOT write int main() function
	vector<int> sol;
	vector<vector<int> > res;
	if(!root) return res;
	queue<TreeNode*> q;
	int count=1;
	q.push(root);
	while(!q.empty())
	{
	int countnew=0;
	while(count)
	{
	TreeNode *tmp=q.front();
	sol.push_back(tmp->val);
	q.pop();
	count--;
	if(tmp->left)
	{
	q.push(tmp->left);
	countnew++;
	}
	if(tmp->right)
	{
	q.push(tmp->right);
	countnew++;
	}
	}
	count=countnew;
	res.push_back(sol);
	sol.clear();
	}
	reverse(res.begin(),res.end());
	return res;
	}
	};

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