Leetcode 树 Populating Next Right Pointers in Each Node II

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Populating Next Right Pointers in Each Node II

Total Accepted: 9695 Total
Submissions: 32965

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,

Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

题意：给定一棵随意二叉树（不一定是perfect binary tree），将它每个节点的next指针都指向该节点右边的节点

思路：bfs

这里不能用dfs了，仅仅能用bfs

bfs遍历将同一层的节点存放在同一个数组里，

然后在遍历每一个数组，将前面的节点和后面的节点connect起来，

最后一个节点和NULL connect起来

须要定义一个新的struct结构，保存指向每一个节点的指针和该节点所在的层

复杂度：时间O(n)， 空间O( n)

相关题目：

Populating Next Right Pointers in Each Node

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:

    struct TreeLinkNodeWithLevel
    {
    	TreeLinkNode *p;
    	int level;
    	TreeLinkNodeWithLevel(TreeLinkNode *pp, int l):p(pp), level(l){}
    };

    void connect(TreeLinkNode *root) {
    	vector<vector<TreeLinkNode *> > nodes;
    	queue<TreeLinkNodeWithLevel> q;
    	q.push(TreeLinkNodeWithLevel(root, 0));
    	while (!q.empty())
    	{
    		TreeLinkNodeWithLevel cur = q.front(); q.pop();
    		if(!cur.p) continue;
    		if(cur.level >= nodes.size()){
    			vector<TreeLinkNode *> temp;
    			temp.push_back(cur.p);
    			nodes.push_back(temp);
    		}else{
    			nodes[cur.level].push_back(cur.p);
    		}
    		TreeLinkNodeWithLevel left(cur.p->left, cur.level + 1);
    		TreeLinkNodeWithLevel right(cur.p->right, cur.level + 1);
    		q.push(left);
    		q.push(right);
    	}
    	for(int i = 0; i < nodes.size(); i++){
    		for(int j = 0; j < nodes[i].size() - 1; j++){
    			nodes[i][j]->next = nodes[i][j + 1];
    		}
    		nodes[i][nodes[i].size() - 1]->next = NULL;
    	}
    }
};