二叉树到堆树的转换 - 陷入无限循环

Question

在解决问题时，我尝试了以下解决方案。不知何故，我的输出陷入了无限循环，没有打印结果或更新的堆树。

给定一棵树，其左右子树都是最小堆，但根节点不维护最小堆属性。您的代码应修改以 Node* n 为根的树，使其成为最小堆。（这意味着你需要满足最小堆属性：一个节点的值可以等于它的一个或两个子节点，但是节点的值不能大于它的任何一个子节点。您不必尝试平衡树或使其成为完整的树。）

#include <iostream>
#include <string>

You have the following class Node already defined.
You cannot change this class definition, so it is
shown here in a comment for your reference only:

class Node {
public:
  int value;
  Node *left, *right;
  Node(int val = 0) { value = val; left = right = nullptr; }
  ~Node() {
    delete left;
    left = nullptr;
    delete right;
    right = nullptr;
  }
};

This function has also previously been defined for you:

void printTreeVertical(const Node* n);

You can use it to print a verbose, vertical diagram of
a tree rooted at n. In this vertical format, a left child
is shown above a right child in the same column. If no
child exists, [null] is displayed.

*/

void downHeap(Node *n) {
    Node *curr = new Node();
    Node *mino = new Node();

  if (n == nullptr ){
    return;
  } else if (n->left->value > n->value & n->right->value > n->value){
    return;
  // } else if (n->left== nullptr & n->right== nullptr) {
  //   return;

  //   } 
  } else {
    // node* curr = new Node(n)
    // n = new Node((std::min(n->left->value, n->right->value));
    // if (n->left->value)
    while(n->left!= nullptr & n->right!= nullptr){
      if (n->left == nullptr){
        mino = n->right;
      } else if (n->right == nullptr) {
        mino = n->left;
      } else {
        mino = (std::min(n->left, n->right));
      }

      std::cout << n->value << std::endl;
      std::cout << mino->value << std::endl;




        if(n->value > mino-> value){
            curr->value = n->value;
            n->value = mino->value;
            mino->value = curr->value;
            std::cout << n->value << std::endl;
            std::cout << mino->value << std::endl;
            downHeap(mino);
          }
        }
        return;
      }
  }

  // Implement downHeap() here.



// You can also use this compact printing function for debugging.
void printTree(Node *n) {
  if (!n) return;
  std::cout << n->value << "(";
  printTree(n->left);
  std::cout << ")(";
  printTree(n->right);
  std::cout << ")";
}

int main() {
  Node *n = new Node(100);
  n->left = new Node(1);
  n->left->left = new Node(3);
  n->right = new Node(2);
  n->right->left = new Node(3);
  n->right->right = new Node(4);
  n->right->right->right = new Node(5);
  std::cout << std::endl << "BEFORE - Vertical printout:" << std::endl;
  printTreeVertical(n);

  downHeap(n);

  std::cout << "Compact printout:" << std::endl;
  printTree(n);
  std::cout << std::endl << " AFTER Vertical printout:" << std::endl;
  printTreeVertical(n);

  delete n;
  n = nullptr;

  return 0;
}

请提出我遗漏的建议。我觉得我把它弄得太复杂了。此外，我没有任何其他功能，如用于将二叉树转换为最小堆的交换功能。我也没有使用数组或向量。因此，如果您能为我提供简单的解决方案，我将不胜感激。

Answer 1

你的主要问题是这行代码：

    mino = (std::min(n->left, n->right));

在这里，当您真正想要比较您所引用的两个对象中的值时，您正在比较两个指针，而 return 是指向具有较小值的对象的指针。即：

mino = (n->left->value < n->right->value) ? n->left : n->right;

同样在这行代码中：

} else if (n->left->value > n->value & n->right->value > n->value){

您可能需要 &&（逻辑与）而不是 &（按位与）。参见 https://www.geeksforgeeks.org/what-are-the-differences-between-bitwise-and-logical-and-operators-in-cc/。

最后，您的代码格式有点不对，所以很难分辨，但看起来 return 语句在 downHeap 函数的 while 循环之外。如果它在循环体之外，则可能导致无限循环。

Answer 2

这是我想到的最简单的解决方案。

Answer 3

尝试此代码以获得完整的解决方案。

void downHeap(Node *n) {
  // check if it is a leaf node
  if (n->left == NULL && n->right == NULL) {
    return;
  }

  if ((n->left != NULL) && (n->right != NULL)) {
    if ((n->value <= n->left->value) && (n->value <= n->right->value)) {
      return;
    } 
  }

  // Node passed in is not a leaf
  // If left is null, we can focus on the right first
  // If the right node has a greater value
  if (n->right != NULL) {
    if ((n->left == NULL) && (n->right->value < n->value)) {
      int rightnodevalue = n->right->value;
      int nodevalue = n->value;

      n->value = rightnodevalue;
      n->right->value = nodevalue;

      if (n->right != NULL) {
        downHeap(n->right);
        return;
      }  
    }
  }


  // Node passed in is not a leaf
  // Left is not null
  // First check if right is null
  // If right is null, we can focus on the left first
  // If the left node has a greater value
  if (n->left != NULL) {
    if ((n->right == NULL) && (n->left->value < n->value)) {
      int leftnodevalue = n->left->value;
      int nodevalue = n->value;

      n->value = leftnodevalue;
      n->left->value = nodevalue;

      if (n->left != NULL) {
        downHeap(n->left);
        return;
      }
    }
  }


  // Node passed in is not a leaf
  // Left is not null
  // Right is not null
  // If left is less than right
  if ((n->left != NULL) && (n->right != NULL)) {
    if (n->left->value < n->right->value) {
      int leftnodevalue = n->left->value;
      int nodevalue = n->value;

      n->value = leftnodevalue;
      n->left->value = nodevalue;

      if (n->left != NULL) {
        downHeap(n->left);
        return;
      }
    }
    else {
      // Proceed to swap the parent and the right node
      // Reference pointers for child's pointers
      int rightnodevalue = n->right->value;
      int nodevalue = n->value;

      n->value = rightnodevalue;
      n->right->value = nodevalue;

      if (n->right != NULL) {
        downHeap(n->right);
        return;
      }
    }
  }

  return;
}

Answer 4

如果您定义一个辅助函数来交换节点值，您可以使您的代码更简洁，并且可以很容易地实现递归。例如：

// DEFINE HELPER FUNCTION
void compareSwap(Node *a, Node *b) {
    int temp;
    if (a->value > b->value) {
        temp = b->value;
        b->value = a->value;
        a->value = temp;
    }
    return;
}

void downHeap(Node *n) {
  Node *compNode;

// 1. n is a leaf
    if (!n->left && !n->right) {
        return;
    }

// 2. n has one left child
  else if (n->left && !n->right) {
    if (n->value > n->left->value) {
      compareSwap(n, n->left);
      downHeap(n->left);
    }
    return;
  }

// 3. n has one right child
  else if (!n->left && n->right) {
    if (n->value > n->right->value) {
      compareSwap(n, n->right);
      downHeap(n->right);
    }
    return;
  }

// 4. n has two children ... (n->left && n->right) 
    else {
    // HEAP IS SATISFIED, RETURN NOTHING
        if ((n->value <= n->left->value) && (n->value <= n->right->value)) {
            return;
        }
    // RIGHT IS LESS THAN n BUT LEFT IS NOT, SWAP RIGHT
    else if (((n->value <= n->left->value) && (n->value > n->right->value))) {
      compareSwap(n, n->right);
      downHeap(n->right);
      return;
    }
    // LEFT IS LESS THAN n BUT RIGHT IS NOT, SWAP LEFT
    else if (((n->value > n->left->value) && (n->value <= n->right->value))) {
      compareSwap(n, n->left);
      downHeap(n->left);
      return;
    }
    // BOTH ARE LESS THAN, SWAP MIN
    else {
      compNode = (n->left->value < n->right->value) ? n->left : n->right; 
      compareSwap(n, compNode);
      downHeap(compNode);
      return;
    }
  }
}

二叉树到堆树的转换 - 陷入无限循环

Binary Tree to Heap Tree Conversion - Stuck in infinite loop

c++

heap

recursion

binary-tree

data-structures