116. Populating Next Right Pointers in Each Node

Given a binary tree

 

    struct TreeLinkNode {

      TreeLinkNode *left;

      TreeLinkNode *right;

      TreeLinkNode *next;

    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

 

Initially, all next pointers are set to NULL.

 

Note:

 

You may only use constant extra space.

You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

         1

       /  \

      2    3

     / \  / \

    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \  / \

    4->5->6->7 -> NULL

 

结题报告：

使用一个指针队列保存树结点，并记录结点索引（从1开始），每层的最后一个结点的next指向null，同层其他结点指向其下一个兄弟结点，每层最后一个结点的索引分别为1，3，7，15，k… 可以看出 k+1是2的幂次方（使用 x&(x-1)==0?判断是否为2的幂次方）。

class Solution { public:     // 28ms     void connect(TreeLinkNode *root) {        queue<TreeLinkNode*>q;        if(root==NULL)return ;        q.push(root);        int k=0;        while(!q.empty()){            TreeLinkNode*p=q.front();            q.pop();            k++;            if(((k+1)&k)!=0&&!q.empty())p->next=q.front();            if(p->left!=NULL)q.push(p->left);            if(p->right!=NULL)q.push(p->right);        }     } };

方法与上类似，只是使用vector来存储树结点。。

class Solution { public:     // 24ms     void connect(TreeLinkNode *root) {        vector<TreeLinkNode*>q;        if(root==NULL)return ;        q.push_back(root);        int k=0;        while(k<q.size()){            TreeLinkNode*p=q[k++];            if(((k+1)&k)!=0&&!q.empty())p->next=q[k];            if(p->left!=NULL)q.push_back(p->left);            if(p->right!=NULL)q.push_back(p->right);        }     } };