Face questions thirty-two: print binary tree from top to bottom

 

要求输出 8,10,6,9,11,5,7
方法一：定义一个队列，把出队的节点的左右子树节点近队，如果左右节点为null则不用进队

 1 void PrintTree(  BinaryTreeNode phead ){  
 2             if(phead==nuull) return ;
 3             Queue<BinaryTreeNode> queue=new Queue();
 4             queue.add(phead);
 5             while( queue.peek()!=null ){
 6                     BinaryTreeNode pNode=queue.poll();  //出队
 7                     System.out.print( pNode.value);
 8 
 9                     //子节点进队
10                     if(pNode.Left!=null)    queue.add(pNode.Left);
11                     if(pNode.Right!=null)   queue.add(pNode.Right);
12             }
13         
14     }

 

Layered print binary increased two variables used to represent a number in the current layer are not printed, a number of the lower layer represents

 Void PrintTree (BinaryTreeNode Pet) {
             IF (Pet == nuull) return ;
            Queue<BinaryTreeNode> queue=new Queue();
            queue.add (Pet);

            int nextPrint = 0;   // the number of nodes in a layer of non-printed 
            int printNum =. 1;   // current number of nodes unprinted layer of 
            the while (! queue.peek () = null ) {
                    BinaryTreeNode pNode=queue.poll();  //出队
                        PrintNum--;
                    System.out.print( pNode.value);

                    // child node into the team 
                    IF (pNode.Left! = Null ) {
                           queue.add(pNode.Left);
                            ++nextPrint;
                     }
                    if(pNode.Right!=null)   {
                            queue.add(pNode.Right);
                            ++nextPrint;
                    }
                    if(PrintNum==0){
                            System.out.println( );
                             PrintNum=nextPrint;;
                             nextPrint;=0;
                    }              
            }
        }

 

Zigzag binary print l example: 8,6,10,9,11,5,7
analysis summary: the print order of the zigzag binary requires two stacks .
If the print odd-numbered layers, the first saved and then save the left node to the right node of another stack;
if the print even-numbered layers, save the first left and right node and then save the stack node to another;

 1 void PrintTree(  BinaryTreeNode phead ){
 2         if(phead==nuull) return ;
 3         
 4         Stack<Integer> s0=new Stack(); //奇数
 5        Stack<Integer> s1=new Stack(); //偶数
 6         
 7         s0.push(phead);
 8         
 9         while( !s1.empty() || !s0.empty() ){
10                 
11                     while(!s0.empty() ){
12                            BinaryTreeNode pNode=s0.pop();
13                             System.out.print( pNode.value );
14                         //先左后右 
15                          if(pNode.Letf)!=null)       s1.push(pNode.Letf) ;
16                          iff(pNode.Right)!=null)         s1.push(pNode.Right );  
17                     }
18                    System.out.println();
19                     while(!s1.empty() ){
20                           BinaryTreeNode pNode=s1.pop();
21                            System.out.print( pNode.value );
22                         //先右后左
23                          if(pNode.Right!=null)        s0.push(pNode.Right );      
24                          if(pNode.Letf)!=null)     s0.push(pNode.Letf) ;                      
25                     }               
26                  System.out.println();
27                 
28         }
29     }