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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
分析：此题是Populating Next Right Pointers in Each Node的升级版，去掉了是perfect binary tree的条件，去掉这个条件带来的影响是我们必须查找节点p孩子的next和下一层的head节点，迭代代码如下：
class Solution { public: void connect(TreeLinkNode *root) { if(root == NULL) return; TreeLinkNode *head = root; while(head){ TreeLinkNode *cur = head; while(cur){ TreeLinkNode *next = NULL, *p; //find next node for(p = cur->next; p && p->left == NULL && p->right == NULL; p = p->next); if(p == NULL) next = NULL; else next = p->left?p->left:p->right; //level link if(cur->left){ cur->left->next = cur->right?cur->right:next; } if(cur->right) cur->right->next = next; cur = cur->next; } //find next head TreeLinkNode *q; for(q = head; q && q->left == NULL && q->right == NULL; q = q->next); if(q == NULL) head = NULL; else head = q->left?q->left:q->right; } } };
上述代码太繁琐，在上面代码的基础上，我们进行简化，实际上在每一层，我们只需维护两个变量，一个next指向下一层的第一个node，一个prev表示同一层的前一个节点。代码如下：
class Solution { public: void connect(TreeLinkNode *root) { while (root) { TreeLinkNode * next = nullptr; // the first node of next level TreeLinkNode * prev = nullptr; // previous node on the same level for (; root; root = root->next) { if (!next) next = root->left ? root->left : root->right; if (root->left) { if (prev) prev->next = root->left; prev = root->left; } if (root->right) { if (prev) prev->next = root->right; prev = root->right; } } root = next; // turn to next level } } };
Leetcode:Populating Next Right Pointers in Each Node II
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原文地址：http://www.cnblogs.com/Kai-Xing/p/4198620.html
