C++ 中二叉树所有叶节点的乘积

给定一棵包含节点的二叉树，任务是找到给定二叉树的所有叶节点的乘积。

叶节点是没有子节点的末端节点。在树中，节点可以作为父节点或子节点，而根节点只能是父节点。因此，具有 NULL 右指针和左指针的节点是叶节点。

输入 

输出 

Leaf nodes are -: 23, 34, 25
Product-: 23*34*25 = 19550

解决方案

  • 输入节点数据

  • 从根节点开始遍历所有节点，并转到左子目录或右子目录进行遍历。

  • 将带有 NULL 右指针和左指针的那些节点存储到临时变量中以查找乘积。

  • 打印包含已乘值的临时变量的值。

算法

Start
Step 1 → create structure of a node and temp, next and head as pointer to a
   structure node
      struct node
         int data
         Create node *left, *right
      End
Step 2 → declare function to insert a node in a tree
   node* new_node(int data)
      Set node* temp = new node()
      Set temp→data = data
      Set temp→left = temp→right = NULL
      return temp
   End
Step 3 → Declare a function to find product of all the leaf nodes
   void leaf(node* root, int &product)
      IF (!root)
         Return
      End
      IF (!root→left && !root→right)
         Set product *= root→data
         Call leaf(root→left, product)
         Call leaf(root→right, product)
Step 4 → In main()
   Create node* root = new_node(10)
   Set root→left = new_node(20)
   Set root→left→left = new_node(30)
   Set int product = 1
   Call leaf(root, product)
   Display product
Stop

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示例

 现场演示

#include <bits/stdc++.h>
using namespace std;
//structure of a node
struct node{
   int data;
   node *left, *right;
};
//function to create a new leaf of a tree
node* new_node(int data){
   node* temp = new node();
   temp→data = data;
   temp→left = temp→right = NULL;
   return temp;
}
//function to find the product of all leaf nodes of a tree
void leaf(node* root, int &product){
   if (!root)
      return;
   if (!root→left && !root->right)
      product *= root→data;
   leaf(root→left, product);
   leaf(root→right, product);
}
int main(){
   node* root = new_node(10);
   root→left = new_node(20);
   root→left→left = new_node(30);
   root→left→right = new_node(40);
   root→right = new_node(50);
   root→right→right = new_node(60);
   root→right→left = new_node(70);
   int product = 1;
   leaf(root, product);
   cout<<"product of a leaf nodes are :"<<product;
   return 0;
}

输出

如果运行以上代码，它将生成以下输出 −

product of a leaf nodes are :5040000

更新于： 13-Aug-2020

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