Question

我需要实现一个void函数，该函数计算二叉树中每个节点的高度并将其存储在每个节点中。我在网上发现了一些本质上是递归的解决方案，但它们返回的都是int。示例包括（https://www.geeksforgeeks.org/write-a-c-program-to-find-the-maximum-depth-or-height-of-a-tree/）。除了不是void函数之外，模型答案之间的区别还在于它不存储每个节点的高度。

这是我尝试的解决方案，但是我似乎无法使代码正常工作，也无法调整模型答案以递归地应用到void函数中。当我在助手代码中运行代码进行测试时，它甚至不显示任何输出。

void computeHeight(Node *n) {
  Node* ltraverser = n;
  Node* rtraverser = n;
  int lheight = 0;
  int rheight =0;

  if (n == NULL) {
   n->height = 0;
  }

  while (ltraverser->left != NULL) {
   ltraverser = ltraverser->left;
   lheight += 1;
  }

  while (rtraverser->right != NULL) {
   rtraverser = rtraverser->right;
   lheight += 1;
  }

  if (lheight > rheight) {
    n->height = lheight;
  }

  else {
    n->height = rheight;
  }

  computeHeight(n->left);
  computeHeight(n->right);
}

供参考：

The starter code below defines a class called "Node" that has two child pointers ("left" , "right") and an integer "height" member variable. There is also a constructor Node() that initializes the children to nullptr and the height to -1.

/*
The height of a node is the number of edges in
its longest chain of descendants.

Implement computeHeight to compute the height
of the subtree rooted at the node n. Note that
this function does not return a value. You should
store the calculated height in that node's own
height member variable. Your function should also
do the same for EVERY node in the subtree rooted
at the current node. (This naturally lends itself
to a recursive solution!)

Assume that the following includes have already been
provided. You should not need any other includes
than these.

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>

You have also 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 height; // to be set by computeHeight()
  Node *left, *right;
  Node() { height = -1; left = right = nullptr; }
  ~Node() {
    delete left;
    left = nullptr;
    delete right;
    right = nullptr;
  }
};
*/

用于测试代码

// This function prints the tree in a nested linear format.
void printTree(const Node *n) {
  if (!n) return;
  std::cout << n->height << "(";
  printTree(n->left);
  std::cout << ")(";
  printTree(n->right);
  std::cout << ")";
}

  Node *n = new Node();
  n->left = new Node();
  n->right = new Node();
  n->right->left = new Node();
  n->right->right = new Node();
  n->right->right->right = new Node();

  computeHeight(n);

  printTree(n);
  std::cout << std::endl << std::endl;
  printTreeVertical(n);

  delete n;
  n = nullptr;

  return 0;
}

Answer 1

代替返回节点高度，只需在左侧和右侧节点上分别调用computeHeight，然后将最大高度存储在节点结构中即可。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>

class Node {
public:
  int height;
  Node *left, *right;
  Node() { height = -1; left = right = nullptr; }
  ~Node() {
    delete left;
    left = nullptr;
    delete right;
    right = nullptr;
  }
};

void computeHeight(Node *node) {

    if (node == nullptr) {
        return;
    }

    computeHeight(node->left);
    computeHeight(node->right);

    int leftHeight = -1;
    int rightHeight = -1;

    if (node->left != nullptr) {
        leftHeight = node->left->height;
    }

    if (node->right != nullptr) {
        rightHeight = node->right->height;
    }

    node->height = std::max(leftHeight, rightHeight) + 1;
}

void printNode(Node *n, int level = 0) {
    if (n == nullptr) {
        return;
    }

    std::cout << std::string(level * 2, ' ') << "Height = " << n->height << "\n";

    printNode(n->left,  level + 1);
    printNode(n->right, level + 1);
}


int main() {
  Node *n = new Node();
  n->left = new Node();
  n->right = new Node();
  n->right->left = new Node();
  n->right->right = new Node();
  n->right->right->right = new Node();


  computeHeight(n);
  printNode(n);
}

Answer 2

您的错误在以下部分，因此，您的程序退出时不显示错误

if (n == NULL) {
   n->height = 0;
}

当n为NULL时；您不应该尝试访问n-> height。如下替换它，您的代码将起作用：

if (n == NULL) {
    return;
}

此外，作为另一个答案，当您要递归计算高度时，不需要while循环，只需使用以下递归公式即可：

Height(n) = 1 + max(Height(n->left), Height(n->right))

此外，出于一致性的原因，通常将NULL子树的高度定义为-1。这样可以使递归公式正常工作。

建议语：为了调试任何程序，一种简单的方法是仅在函数调用和/或某些行之前和之后打印消息。通过检查哪些消息未打印，您可以快速查明导致问题的功能/行，然后进行调查。

实现递归的虚函数（二进制搜索树的查找高度）

2 个答案: