在C语言中，打印给定层级的叶节点

The task involves printing leaf nodes of a binary tree at given level k which is specified by the user.

Leaf nodes are the end nodes whose left and right pointer is NULL which means that particular node is not a parent node.

Example

的中文翻译为：

示例

Input : 11 22 33 66 44 88 77
Output : 88 77

在这里，k代表需要打印的树的层级。这里使用的方法是遍历每个节点，并检查节点是否有任何指针。即使只有一个指针，表示左侧或右侧或两者都有，那个特定的节点也不能是叶节点。

使用层次遍历技术递归遍历每个节点，从左侧开始，然后是根节点，最后是右侧。

下面的代码展示了给定算法的C语言实现

算法

START
   Step 1 -> create node variable of type structure
      Declare int data
      Declare pointer of type node using *left, *right
   Step 2 -> create function for inserting node with parameter as new_data
      Declare temp variable of node using malloc
      Set temp->data = new_data
      Set temp->left = temp->right = NULL
      return temp
   Step 3 -> declare Function void leaf(struct node* root, int level)
      IF root = NULL
         Exit
      End
      IF level = 1
         IF root->left == NULL && root->right == NULL
            Print root->data
         End
      End
      ELSE IF level>1
         Call leaf(root->left, level - 1)
         Call leaf(root->right, level - 1)
      End
   Step 4-> In main()
      Set level = 4
      Call New passing value user want to insert as struct node* root = New(1)
      Call leaf(root,level)
STOP

Example

的中文翻译为：

示例

include<stdio.h>
#include<stdlib.h>
//structre of a node defined
struct node {
   struct node* left;
   struct node* right;
   int data;
};
//structure to create a new node
struct node* New(int data) {
   struct node* temp = (struct node*)malloc(sizeof(struct node));
   temp->data = data;
   temp->left = NULL;
   temp->right = NULL;
   return temp;
}
//function to found leaf node
void leaf(struct node* root, int level) {
   if (root == NULL)
      return;
   if (level == 1) {
      if (root->left == NULL && root->right == NULL)
      printf("%d</p><p>",root->data);
   } else if (level > 1) {
      leaf(root->left, level - 1);
      leaf(root->right, level - 1);
   }
}
int main() {
   printf("leaf nodes are: ");
   struct node* root = New(11);
   root->left = New(22);
   root->right = New(33);
   root->left->left = New(66);
   root->right->right = New(44);
   root->left->left->left = New(88);
   root->left->left->right = New(77);
   int level = 4;
   leaf(root, level);
   return 0;
}

输出

如果我们运行上述程序，它将生成以下输出。

leaf nodes are: 88 77

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