Q: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

A: recursive.

树的问题优先考虑用递归解决。

void combine(TreeLinkNode* leftRoot,TreeLinkNode* rightRoot)    {        if(!leftRoot&&!rightRoot)            return;        leftRoot->next = rightRoot;        TreeLinkNode *it1,*it2;        for(it1=leftRoot->right,it2=rightRoot->left;it1&&it2;it1 = it1->right,it2 = it2->left)            it1->next = it2;       }        void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(!root)            return;        root->next = NULL;        connect(root->left);        connect(root->right);        combine(root->left,root->right);            }

　　

更简单的方法。。之前没想到，按原先的想法话，有些结点重复遍历。以下的方法，将tree一层一层来看待，一层一层链接，结点只遍历一次。

void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(!root||(!root->left&&!root->right))            return;        root->left->next = root->right;        if(root->next)            root->right->next = root->next->left;        connect(root->left);        connect(root->right);            }