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LeetCode OJ 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
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解答
利用已经连好的next指针进行广度优先遍历,注意是每次处理下一层,所以下一层为NULL的时候就不需要处理了。
/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * struct TreeLinkNode *left, *right, *next; * }; * */ void connect(struct TreeLinkNode *root) { struct TreeLinkNode *head, *next_head, *pNode, *pre_node; if(NULL == root){ return; } root->next = NULL; next_head = root; while(1){ head = next_head; next_head = next_head->left; if(NULL == next_head){ break; } for(pNode = head; pNode != NULL; pNode = pNode->next){ if(pNode != head){ pre_node->next = pNode->left; } pNode->left->next = pNode->right; pre_node = pNode->right; } pre_node->next = NULL; } }
LeetCode OJ 116. Populating Next Right Pointers in Each Node
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