Question

我试图获取二叉树的镜像，但是由于对指针的了解较少，我很难调试代码。可以帮助我调试代码......

算法：

使用级别订单travesal（从右到左）遍历树
然后在弹出前节点时创建镜像树

See my code here.

我在做什么？

将root插入队列，然后运行while（！queue.empty（））。
弹出前面元素并将其插入镜像树
然后，按前 - >右，然后按前 - >左。

我知道这是一个非常微不足道的问题，但仅仅是因为我清楚地了解了我在这里发布它的指针。我也无法理解我的编译器在说什么。在我尝试编译时，她是一个快照。任何帮助将不胜感激。谢谢!!!

Answer 1

我已经更正了代码。请查看是否有帮助

#include<bits/stdc++.h>
using namespace std;

struct treenode{
  int data;
  struct treenode* left , *right;
};


struct treenode* createNode(int data)
{
  struct treenode* node = (struct treenode*)malloc(sizeof(struct treenode));
  node->data = data;
  node->right = NULL;
  node->left = NULL;
  return node;
}


//Function for inserting node for creating mirror of binary tree
void insert(struct treenode **root2,int data)
{
  struct treenode* temp;
  if(*root2 == NULL)
  {
    *root2 = createNode(data);
    return;
  }
  else{
    queue<treenode*> q;
    q.push(*root2);
    while(!q.empty())
    {
      temp = q.front();
      q.pop();
      if(temp->left)
      q.push(temp->left);
      else{
        temp->left = createNode(data);
        return;
      }
      if(temp->right)
      {
        q.push(temp->right);
      }
      else{
        temp -> right = createNode(data);
        return;
      }
    }
  }
}

void mirror(struct treenode* node) 
{
  if (node==NULL) 
    return;  
  else
  {
    struct treenode* temp;

    /* do the subtrees */
    mirror(node->left);
    mirror(node->right);

    /* swap the pointers in this node */
    temp        = node->left;
    node->left  = node->right;
    node->right = temp;
  }
} 

//For printing the tree
void level_order(struct treenode** root)
{
  if(root==NULL) return;
  queue<treenode*> q;
  q.push(*root);
  while(!q.empty())
  {
    struct treenode* temp = q.front();
    q.pop();
    cout << temp->data << " ";
    if(temp->left)
    q.push(temp->left);
    if(temp->right)
    q.push(temp->right);
  }
}

//Driver Function
int main()
{
  struct treenode* root2 = NULL;
  struct treenode* root = createNode(1);
  root->left = createNode(3);
  root->right = createNode(2);
  root->right->left = createNode(5);
  root->right->right = createNode(4);
  level_order(&root);
  mirror(root);
  printf("\n");
  level_order(&root);
  return 0;
}

输出：

     1 3 2 5 4

     1 2 3 4 5

Answer 2

//Function for creating Binary Tree
//Assuming *root for original tree and *root2 for mirror tree
//are declared globally
void insert(int data)
{
struct treenode* temp;
if(root2 == NULL)
{
root2 = createNode(data);
return;
}
else{
queue<treenode*> q;
q.push(root2);
while(!q.empty())
{
  temp = q.front();
  q.pop();
  if(temp->left)
  q.push(temp->left);
  else{
    temp->left = createNode(data);
    return;
  }
  if(temp->right)
  {
    q.push(temp->right);
  }
  else{
    temp -> right = createNode(data);
    return;
  }
 }
 }
 }
//Iterative version for Mirror of a Binary Tree 
void mirrorBinaryTree()
{
struct treenode *front;
queue<treenode*> q;
q.push(root);
while(!q.empty())
{
  front = q.front();
  insert(front->data);
  q.pop();
  if(front->right)
  q.push(front->right);
  if(front->left)
  q.push(front->left);
}
}

Answer 3

我认为如果你递归编码会更容易。一些如：

Node * mirror(Node * root)
{
  if (root == nullptr)
    return nullptr;

  Node * r = new Node; // here we will copy the root
  r->key = root->key; // or whatever the root contains
  r->left = mirror(root->right);
  r->right = mirror(root->left);

  return r;
}

使用STL队列的二叉树的镜像

3 个答案: