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
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
二叉树的结构体里增加了next指针,编写程序,将二叉树里每个节点的next指针指向它右边的节点。
思路,对二叉树进行层序遍历,使用previous指针记录上一个节点,遍历到当前节点时将previous的next指向当前节点,并将previous改为当前节点进行下一次循环;使用num1,num2来记录每层的节点个数,来区分每层的界限,更具体的逻辑可以参考程序。
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==NULL) return; queue<TreeLinkNode*> qt; TreeLinkNode *previous=NULL; int num1=1,num2=0; qt.push(root); while(!qt.empty()) { while(num1>0) { root=qt.front(); qt.pop(); num1--; if(root->left!=NULL) { qt.push(root->left); num2++; } if(root->right!=NULL) { qt.push(root->right); num2++; } if(previous!=NULL) previous->next=root; previous=root; } previous->next=NULL; previous=NULL; num1=num2; num2=0; } } };
leetcode 刷题之路 81 Populating Next Right Pointers in Each Node
时间: 2024-11-08 03:58:57