題記：寫這篇博客要主是加深自己對按層空間復雜度的認識和總結實現算法時的一些驗經和訓教，如果有錯誤請指出，萬分感謝。

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

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

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Note:? You may only use constant extra space.

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?

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For example,
Given the following binary tree,

    1

       /  \

      2    3

     / \    \

    4   5    7
  

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?

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After calling your function, the tree should look like:

    1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \    \

    4-> 5 -> 7 -> NULL
  

????碼好一次性通過~~ yeah~

????路思是很明顯的，一層一層的找，利用前當層的next來figure out下一層的next

????第一次做的時候用了queue來實現的按層遍歷，此次才看到了 O(1)的空間復雜度。

????從新寫了一下，為了看起來路思清楚點，碼代寫的略長。

    /**

 * 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:

    TreeLinkNode* myNextLine(TreeLinkNode *cur){

        if(cur -> left) return cur -> left;

        if(cur -> right) return cur -> right;

        

        while(cur -> next){

            cur = cur -> next;

            if(cur -> left) return cur -> left;

            if(cur -> right) return cur -> right;

        }

        

        return NULL;

    }



    TreeLinkNode* myleftnext(TreeLinkNode *cur){

        if(cur -> right) return cur -> right;

        

        while(cur -> next){

            cur = cur -> next;

            if(cur -> left) return cur -> left;

            if(cur -> right) return cur -> right;

        }

        

        return NULL;

    }

    

    TreeLinkNode* myrightnext(TreeLinkNode *cur){

        while(cur -> next){

            cur = cur -> next;

            if(cur -> left) return cur -> left;

            if(cur -> right) return cur -> right;

        }

        

        return NULL;

    }

    



    void connect(TreeLinkNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        if(!root) return;

        

        TreeLinkNode *cur = root;

        TreeLinkNode *nextLine = myNextLine(cur);

        

        while(nextLine){

            while(cur){

                if(cur -> left){

                    cur -> left -> next = myleftnext(cur);

                }

                if(cur -> right){

                    cur -> right -> next = myrightnext(cur);

                }

                

                

                cur = cur -> next;

            }

            cur = nextLine;

            nextLine = myNextLine(cur);

        }

        

    }

};
  

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按層空間復雜度Populating Next Right Pointers in Each Node II