"Leetcode-Interview algorithm classic-java Implementation" "" 117-populating Next right pointers in each Node (binary tree link to starboard pointer ii) "

"117-populating Next right pointers in each Node (binary tree link to starboard pointer ii)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

Follow up to problem "populating Next right pointers in each Node".
　　What if the given tree could is any binary tree? Would your previous solution still work?
　　Note:
Constant extra space.
For example,
Given The following binary tree,

         1       /        2    3     \        4   5    7

After calling your function, the tree is should look like:

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

Main Topic

Given a binary tree, there is a next pointer that links each of their layers. Only constant extra space is used, and the tree is an arbitrary two- fork tree.

Thinking of solving problems

String Each node of the tree together with next. Each layer will also form a single linked list. And each layer of the linked list head, is, the root of the left child, left child, left child. Using a double loop, the outer loop, along the root of the left child, has been downward. The inner loop is responsible for stringing the nodes of the next layer together. That is, put your right child on the left child's next, and the right child, you can find the right neighbor through your next pointer.

Code Implementation

Tree Node class

publicclass TreeLinkNode {    TreeLinkNode left;    TreeLinkNode right;    TreeLinkNode next;}

Algorithm implementation class, using constant space

 Public classSolution { Publicvoid Connect (Treelinknode root) {Treelinknode queue = root; Treelinknode level =NewTreelinknode (0); while(Queue! =NULL) {level.Next=NULL; Treelinknode current = level; while(Queue! =NULL) {if(Queue. Left!=NULL) {Current.Next= Queue. Left; Current = current.Next; }if(Queue. Right!=NULL) {Current.Next= Queue. Right; Current = current.Next; } queue = queue.Next; } queue = level.Next; }    }}

Algorithm implementation class, using very much space

ImportJava.Util.Iterator;ImportJava.Util.LinkedList;ImportJava.Util.List; PublicClass Solution { Public voidConnect (Treelinknode root) {if(Root!= NULL) {//Save node            List<Treelinknode> List = NewLinkedList<>();//The previous node of the currently processed nodeTreelinknode prev= NULL;//node currently being processedTreelinknode node;///number of nodes remaining in the current layerint Curr= 1;//Record the number of elements in the next layerint Next= 0;//The root node is in the queue.            List.Add (root);//Queue non-empty             while(List.Size ()> 0) {//Delete team first elementNode= List.Remove0);///The number of current layer remaining is reducedCurr--;//Left dial hand tree non-empty, left Dial hand node queue                if(node.Left!= NULL) {List.Add (node.left); Next++; }//Right sub-tree non-empty, right sub-node queue                if(node.Right!= NULL) {List.Add (node.right); Next++; }//If the current layer is finished processing                if(Curr== 0) {//thread the elements on the next layerIterator<Treelinknode>Iterable= List.Iterator ();if(iterable.Hasnext ()) {prev=Iterable.Next (); while(iterable.Hasnext ()) {node=Iterable.Next (); Prev.Next=Node Prev=Node }                    }//Update the remaining node points in the current layerCurr=Next//re-count lower node pointsNext= 0; }            }        }    }}

Evaluation Results

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Use constant space

Use a very good amount of space

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"Leetcode-Interview algorithm classic-java Implementation" "" 117-populating Next right pointers in each Node (binary tree link to starboard pointer ii) "