LeetCode Populating Next Right Pointers in Each Node
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题目
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 toNULL
.
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
输出每个节点右边的节点,bfs层序遍历即可。
代码:
/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public: void connect(TreeLinkNode *root) { TreeLinkNode *pos,*next_level=root,*next_head=root;//当前处理位置,下一层处理到位置,下一层头 while(next_head!=NULL)//层序遍历 { pos=next_head; next_level=NULL; next_head=NULL; while(pos!=NULL)//有未处理的层 {if(next_level==NULL)//处理左节点{next_level=pos->left;next_head=next_level;}else{next_level->next=pos->left;next_level=next_level->next;}if(pos->right==NULL)//处理右节点break;next_level->next=pos->right;next_level=next_level->next;pos=pos->next; } } }};
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