[LeetCode] Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \    \    4-> 5 -> 7 -> NULL

解题思想：

就是要找到每一个开始的节点，用BFS,把每一层走完。

递推思想 依然不需要改变，依然是依据当前层的next 指针，设置下一层的 next 指针。只是找结点麻烦些，我们定义了两个函数，findNextNodeNextLev用来找(n+1)层的下一个节点，findStartNodeNextLev来找这一层的第一个节点。通过调用，findNextNodeNextLev来找，找到了返回。

class Solution {public:    void connect(TreeLinkNode *root)     {if (NULL==root){return;}TreeLinkNode *start;TreeLinkNode *curNode;    TreeLinkNode *nextNode;while (root!=NULL){    start = findStartNodeNextLev(root);    curNode=start;    nextNode=findNextNodeNextLev(root,start); //nextLev First，nexNode    while(nextNode!=NULL)    {      curNode->next=nextNode;      curNode=nextNode;              nextNode=findNextNodeNextLev(root,curNode);    }root=start;}            }    private:    TreeLinkNode* findNextNodeNextLev(TreeLinkNode* &cur,TreeLinkNode* curNextLev)    {        if(cur->left==curNextLev&&cur->right!=NULL)        {            return cur->right;        }        else        {            while(cur->next!=NULL)            {                cur=cur->next;                if(cur->left!=NULL&&cur->left!=curNextLev) return cur->left;                if(cur->right!=NULL&&cur->right!=curNextLev) return cur->right;            }        }        return NULL;    }    TreeLinkNode* findStartNodeNextLev(TreeLinkNode* node)    {        if(NULL == node) return NULL;        if(node->left!=NULL)return node->left;        return findNextNodeNextLev(node,node->left);    }};