【Populating Next Right Pointers in Each Node II】cpp

题目：

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

代码：

/**
 * 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) return;
            deque<TreeLinkNode *> curr, next;
            curr.push_back(root);
            while ( !curr.empty() )
            {
                TreeLinkNode dummy(-1);
                TreeLinkNode *pre = &dummy;
                while ( !curr.empty() )
                {
                    TreeLinkNode *tmp = curr.front(); curr.pop_front();
                    pre->next = tmp;
                    if (tmp->left) next.push_back(tmp->left);
                    if (tmp->right) next.push_back(tmp->right);
                    pre = tmp;
                }
                pre->next = NULL;
                std::swap(curr, next);
            }
    }
};

tips：

广搜思路（有些违规，因为不是const extra space）

每一层设立一个虚拟头结点，出队的同时pre->next = tmp

=====================================

学习了另外一种思路，可以不用队列的数据结构，这样就符合const extra space的条件了。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) {
            TreeLinkNode *pre;
            TreeLinkNode *curr;
            TreeLinkNode *next_first; // store next level's first not null node
            curr = root;
            while ( curr ){
                // move to next tree level
                TreeLinkNode dummy(-1);
                pre = &dummy;
                next_first = NULL;
                // connect the curr level
                // record the first not null left or right child of the curr level as for the first node of next level
                while ( curr ){
                    if ( !next_first ){
                        next_first = curr->left ? curr->left : curr->right;
                    }
                    if ( curr->left ){
                        pre->next = curr->left;
                        pre = pre->next;
                    }
                    if ( curr->right ){
                        pre->next = curr->right;
                        pre = pre->next;
                    }
                    curr = curr->next;
                }
                curr = next_first;
            }
    }
};

tips：

这套代码的大体思路是数学归纳法

1. root节点的root->next按照题意是NULL

2. 处理root的left节点和right节点之间的next关系

...如果知道了第n-1层的node之间的next关系，则可以得到第n层node节点之间的next关系...

按照这套思路，就可以写出上述的代码。

这里有两个细节需要注意：

a. next_first不一定是left还是right，这个要注意，只要next_first一直为NULL就要一直找下去。

b. 设立一个dummy虚node节点，令pre指向dummy，可以不用判断pre为NULL的情况，简化判断条件。

另，回想为什么level order traversal的时候必须用到队列，可能就是因为没有每一层之间的next关系。

==========================================

第二次过这道题，直接看常数空间复杂度的思路，重新写了一遍AC。

/**
 * 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) {
            TreeLinkNode* curr = root;
            while ( curr )
            {
                TreeLinkNode* pre = new TreeLinkNode(0);
                TreeLinkNode* next_level_head = NULL;
                while ( curr )
                {
                    if ( !next_level_head )
                    {
                        next_level_head = curr->left ? curr->left : curr->right;
                    }
                    if ( curr->left )
                    {
                        pre->next = curr->left;
                        pre = pre->next;
                    }
                    if ( curr->right )
                    {
                        pre->next = curr->right;
                        pre = pre->next;
                    }
                    curr = curr->next;
                }
                curr = next_level_head;
            }
    }
};