leetcode--Populating Next Right Pointers in Each Node
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Problem Description:
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
思路分析:
利用队列记录每个节点的指针,实现层次遍历,遍历的同时在每一层用一个计数器记录该层的节点个数,依次对next指针赋值。
代码如下:
/** * 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) { if(!root) return; int i=0,levellength; queue<TreeLinkNode *> tree; tree.push(root); levellength=1; while(!tree.empty()) { while(i<levellength) { TreeLinkNode *p=tree.front(); if(p->left) tree.push(p->left); if(p->right) tree.push(p->right); tree.pop(); if(i!=levellength-1) p->next=tree.front(); else p->next=NULL; i++; } levellength=tree.size(); i=0; } }};
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