从顶层开始逐层打印二叉树节点数据

从顶层开始逐层打印二叉树节点数据

               10

              /   \

           6       14

         /  \      /   \

       4    8   12   16

逐层打印的结果为  10  6  14  4  8 12  16

采用队列的方式 ，先把根节点存入队列，然后再出列，输出这个数据。如果有左子树，把左子树根节点入列，如果右子树不为空，把右子树的根节点入列，然后如果队列不为空，则出列。直到队列为空。

 

void  print(TreeNode * root)

{

       queue<TreeNode> tree; //声明一个树节点的队列

       TreeNode tmp = NULL;

        tree.push(root);  //首先把根节点入列

        while( ! tree.empty())

        {

               tmp = tree.front();

               tree.pop();

               (*visit)(tmp);

               if(tmp->left != NULL)

                {

                     tree.push(tmp->left);

                 }

                if(tmp->right != NULL)

                {

                     tree.push(tmp->right);

               }

         }

 

    

}

 

void visit(TreeNode *node)

{

printf("%d",node->data);

}

 

#include <iostream>//#include <deque>#include <queue>using namespace std;typedef struct BtreeNode{int data;struct BtreeNode *left ,*right;}node;void create(node *&r ,int m){if(r==NULL){node *t =new node();t->data = m;t->left =NULL;        t->right =NULL;r = t;}else{       if(m>r->data)   {create(r->right ,m);   }   if(m<r->data)   {create(r->left ,m);   }}}void print(node *r){queue<node*>tmp;tmp.push(r);while(!tmp.empty()){    node *p = tmp.front();cout<<p->data<<endl;tmp.pop();if(p->left){tmp.push(p->left);}if(p->right){tmp.push(p->right);}}} jingxiang(node *&r){  if(!r)  {  return;  }  node *p = r;  p = r->left;  r->left = r->right;  r->right = p;  if(r->left)  {  jingxiang(r->left);  }  if(r->right)  {  jingxiang(r->right);  }}void main(){node *root=NULL;create(root , 10);create(root , 6);create(root , 4);create(root , 8);create(root , 14);create(root , 12);create(root , 16);print(root);jingxiang(root);print(root);}