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：给定一棵任意二叉树（不一定是perfect binary tree），将它每一个节点的next指针都指向该节点右边的节点

思路：bfs
这里不能用dfs了，只能用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
需要定义一个新的struct结构，保存指向每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;    }    }};