标签：style blog color io strong for ar cti
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:
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
思路：自顶向下链接二叉树的每一层，当处理到第i层时，第i-1层的所有节点已经构成一个单向链表。依次遍历第i-1层便可顺利完成对第i层的链接。因为去掉了输入是完全二叉树的限定，因此内层循环比完全二叉树的会稍微复杂一些。
1 class Solution { 2 public: 3 void connect( TreeLinkNode *root ) { 4 TreeLinkNode *head = root; 5 while( head ) { 6 TreeLinkNode *newHead = 0, *prevNode = 0; 7 while( head ) { 8 if( head->left ) { 9 if( !prevNode ) { 10 newHead = prevNode = head->left; 11 } else { 12 prevNode->next = head->left; 13 prevNode = prevNode->next; 14 } 15 } 16 if( head->right ) { 17 if( !prevNode ) { 18 newHead = prevNode = head->right; 19 } else { 20 prevNode->next = head->right; 21 prevNode = prevNode->next; 22 } 23 } 24 head = head->next; 25 } 26 head = newHead; 27 } 28 return; 29 } 30 };
Populating Next Right Pointers in Each Node II,布布扣,bubuko.com
Populating Next Right Pointers in Each Node II
标签：style blog color io strong for ar cti
原文地址：http://www.cnblogs.com/moderate-fish/p/3914121.html
