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 to NULL
.
Initially, all next pointers are set to NULL
.
Note:
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
解答:
要利用好“完美二叉树”这一前提条件,即每一层的节点个数为2的整数次幂,且有左子树时必有右子树。每到整数次幂的节点,next 就指向 NULL.
而且,处理1,然后处理2、3,然后处理4、5、6、7……顺序的访问方式,是不是很像数据结构:队列。
/** * 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; if(root->left == NULL && root->right == NULL) return; queue<TreeLinkNode*> s; s.push(root); int i = 0; int j = 1; TreeLinkNode *tmp; while(!s.empty()) { tmp = s.front(); s.pop(); i++; if(i == j) { i = 0; j = j*2; tmp->next = NULL; } else { tmp->next = s.front(); } if(tmp->left) { s.push(tmp->left); s.push(tmp->right); } } } };
【LeetCode从零单刷】Populating Next Right Pointers in Each Node
原文地址:http://blog.csdn.net/ironyoung/article/details/45015801