[LeetCode] 116. Populating Next Right Pointers in Each Node 解决思路

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

问题： 给定一个二叉树，将树元素的 *next 指向该元素在树结构中的水平右边节点。

这是广度遍历的一个应用。可以借组队列结构实现广度遍历，求解题目。

思路：

• 当队列中元素恰好是树种某一行的全部元素时，则给队列中每个节点的 *next 赋值为 列表中对应的下一个节点，其中特别地，最后一个元素*next 为 NULL。
• 将队列中的元素全部弹出，并依次塞进他们的子节点，此时，队列中的元素恰好是树种下一行的全部元素，继续上一步操作。

将根节点塞进队列，即实现了上面思路的初始化。

 1     void connect(TreeLinkNode *root) {
 2
 3         if(root == NULL){
 4             return;
 5         }
 6
 7         list<TreeLinkNode*> queue;
 8
 9         queue.push_back(root);
10
11         while(queue.size() > 0 ){
12
13             // assign value to the next point of the node in queue.
14             list<TreeLinkNode*>::iterator q_iter;
15             for( q_iter = queue.begin() ; std::next(q_iter,1) != queue.end(); q_iter++){
16                 (*q_iter)->next = *std::next(q_iter,1);
17             }
18
19             // pop each node in the current row in the tree structure, and push the left and right childrens of them into queue.
20             while(queue.front()->next != NULL){
21                 TreeLinkNode* node = queue.front();
22                 queue.pop_front();
23
24                 if(node->left != NULL){
25                     queue.push_back(node->left);
26                 }
27
28                 if(node->right != NULL){
29                     queue.push_back(node->right);
30                 }
31             }
32
33             TreeLinkNode* node = queue.front();
34             queue.pop_front();
35
36             if(node->left != NULL){
37                 queue.push_back(node->left);
38             }
39
40             if(node->right != NULL){
41                 queue.push_back(node->right);
42             }
43         }
44     }