[LeetCode] Binary Tree Level Order Traversal 2

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],
]

head.h里代码实现如下：

#include<vector>
#include<queue>
using namespace std;
#define NULL 0

struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x):val(x),left(NULL),right(NULL) {}
};

struct Node
{
	TreeNode *node;
	int level;
	Node(){}
	Node(TreeNode *n,int l):node(n),level(l){}
};

class Solution
{
private:
	vector<vector<int> > ret;
	vector<vector<int> > retReverse;
public:
	vector<vector<int> > leverOrder(TreeNode *root)
	{
		ret.clear();

		if(root == NULL)
			return ret;

		queue<Node> q;
		q.push (Node(root,0));

		vector<int> a;
		int curLever = -1;

		while(!q.empty() )
		{
			Node node = q.front();
			if(node.node->left)
				q.push(Node(node.node->left,node.level+1));
			if(node.node->right)
				q.push(Node(node.node->right,node.level+1));

			if(curLever != node.level)
			{
				if(curLever != -1)
					ret.push_back(a);
				curLever = node.level ;
				a.clear();
				a.push_back(node.node->val);
			}
			else
				a.push_back(node.node->val);

			q.pop();
		}

		ret.push_back(a);

		retReverse.clear();

		for(int i=ret.size()-1;i>=0;i--)
			retReverse.push_back(ret[i]);

		return retReverse;

	}
};

main.cpp里代码实现如下：

#include"head.h"
#include<iostream>

int main()
{
 TreeNode *root;
 root=new TreeNode(3);
 root->left = new TreeNode(9);
 root->right =  new TreeNode(20);
 root->right->left =  new TreeNode(15);
 root->right->right =  new TreeNode(7);
 Solution sol;
 for(vector<vector<int> >::size_type ix=0;ix!=sol.leverOrder(root).size();++ix)
  {for(vector<int>::size_type i=0;i!=sol.leverOrder(root)[ix].size();++i)
       cout<<sol.leverOrder(root)[ix][i]<<" ";
      cout<<endl;}
 delete  root->right->right;
 delete  root->right->left;
 delete  root->right;
 delete  root->left;
 delete  root;
 return 0;
}