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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
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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
这道题实际上是树的层序遍历的应用,可以参考之前的博客Binary Tree Level Order Traversal 二叉树层序遍历,既然是遍历,就有递归和非递归两种方法,最好两种方法都要掌握,都要会写。下面先来看递归的解法,由于是完全二叉树,所以若节点的左子结点存在的话,其右子节点必定存在,所以左子结点的next指针可以直接指向其右子节点,对于其右子节点的处理方法是,判断其父节点的next是否为空,若不为空,则指向其next指针指向的节点的左子结点,若为空则指向NULL,代码如下:
// Recursion class Solution { public: void connect(TreeLinkNode *root) { if (!root) return; if (root->left) root->left->next = root->right; if (root->right) root->right->next = root->next? root->next->left : NULL; connect(root->left); connect(root->right); } };
对于非递归的解法要稍微复杂一点,但也不算特别复杂,需要用到queue来辅助,由于是层序遍历,每层的节点都按顺序加入queue中,而每当从queue中取出一个元素时,将其next指针指向queue中下一个节点即可。代码如下:
// Non-recursion class Solution { public: void connect(TreeLinkNode *root) { if (!root) return; queue<TreeLinkNode*> q; q.push(root); q.push(NULL); while (true) { TreeLinkNode *cur = q.front(); q.pop(); if (cur) { cur->next = q.front(); if (cur->left) q.push(cur->left); if (cur->right) q.push(cur->right); } else { if (q.size() == 0 || q.front() == NULL) return; q.push(NULL); } } } };
[LeetCode] Populating Next Right Pointers in Each Node 每个节点的右向指针
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原文地址:http://www.cnblogs.com/grandyang/p/4288151.html