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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:
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：
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 }
[LeetCode] 116. Populating Next Right Pointers in Each Node 解决思路
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原文地址：http://www.cnblogs.com/TonyYPZhang/p/5118051.html
