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Populating Next Right Pointers in Each Node I II 均AC

2018年04月29日 ⁄ 综合 ⁄ 共 1302字 ⁄ 字号 评论关闭

感觉写的真啰嗦

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

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

queue<vector<TreeLinkNode*> > q;

class Solution {
public:
    void connect(TreeLinkNode *root) 
	{
        bool first = true;
        if(!root)
      	 	return;
        vector<TreeLinkNode*> v1;
        v1.push_back(root);
        q.push(v1);
        TreeLinkNode *pre = NULL;

        while(!q.empty())
        {
        	vector<TreeLinkNode*> t = q.front();
        	q.pop();
        	
        	vector<TreeLinkNode*>::iterator iter;
        	vector<TreeLinkNode*> vv;
        	first = true;
        	for(iter = t.begin(); iter != t.end(); ++iter)
        	{
        		if(first)
       			{
   				   	pre = *iter;
			    	first = false;
					if(pre->left)
						vv.push_back(pre->left);
					if(pre->right)
        				vv.push_back(pre->right);
					continue;   	
		       	}

	        	TreeLinkNode *p = *iter;
	        	pre->next = p;
	        	p->next = NULL;
	        	pre = p;
	        	
	        	if(p->left)
	        		vv.push_back(p->left);
        		if(p->right)
        			vv.push_back(p->right);
	        }
	        pre->next = NULL;
			if (vv.size() > 0)
			{
				q.push(vv);
			}
			else
		 		break;
        }
    }
};

int main(void)
{
	Solution s;
	
	TreeLinkNode *root = new TreeLinkNode(1);
	TreeLinkNode *node1 = new TreeLinkNode(2); root->left = node1;
	TreeLinkNode *node2 = new TreeLinkNode(3); root->right = node2;
	TreeLinkNode *node3 = new TreeLinkNode(4); node1->left = node3;
	TreeLinkNode *node4 = new TreeLinkNode(5); node1->right = node4;
	TreeLinkNode *node5 = new TreeLinkNode(6); node2->left = node5;
	TreeLinkNode *node6 = new TreeLinkNode(7); node2->right = node6;
	
	s.connect(root);
	cout << node4->next->val << endl;
	
	return 0;
}

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