数据结构树二叉树计算节点

Algorithm:

算法：

One of the popular traversal techniques to solve this kind of problems is level order tree traversal (Read: Level Order Traversal on a Binary Tree) where we use the concept of BFS.

解决此类问题的一种流行的遍历技术是级别顺序树遍历(阅读： 二叉树上的级别顺序遍历 )，我们使用BFS的概念。

The basic idea to solve the problem is:

解决问题的基本思路是：

1. Do a level order traversal and check whether the processed node has its left and right child both NULL.

   进行级别顺序遍历，并检查已处理节点的左，右子节点是否均为NULL 。

2. If the processed node has its left and right child both NULL, it’s a leaf node.

   如果处理后的节点的左右子节点均为NULL，则为叶节点。

3. Increase leaf node count

   增加叶节点数

Pseudocode:

伪代码：

struct BT{              // tree node type
int data;           //value
struct BT *left;    //pointer to left child
struct BT *right;   //pointer to right child
};
int noofleafnodes(struct BT *root){ // root of the tree
// BT refers to node of tree (datatype for the node);
struct BT *temp;                
struct Queue *q=Creat_queue();  //creating a queue
int count=0;
if(!root)   //root is null
return;
//**using level oder search**
EnQueue(q,root); // Enqueue the root in the queue
while(!emptyQueue(q)){
temp=DeQueue(q);  //Dequeue
// processing the node and checking whether both child nodes are NULL or not 
if(!temp->left && !temp->right) 
//increase no of leaves count if both child nodes are NULL (ensures it's a leaf  node)
count++;           
else{
if(temp->left)
EnQueue(q,temp->left); // if left child exists EnQueue
if(temp->right)
EnQueue(q,temp->right); // if right child exists EnQueue
}
}
DeleteQueue(q);
return count; // return no of leaf nodes
}

Image source: wikipedia

图片来源：Wikipedia

Example:

例：

The leaf nodes in the above tree is 2, 5, 11, 4

上面树中的叶节点是2、5、11、4

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
class tree{    // tree node is defined
public:
int data;
tree *left;
tree *right;
};
int noofleafnodes( tree *root){
queue<tree*> q;  // using stl
tree* temp;
int count=0;
q.push(root);
while(!q.empty()){
temp=q.front();
q.pop();
//process node and check wheather leaf node or not
if(!temp->left && !temp->right){            
count++;
cout<<temp->data<<" ";
}
if(temp->left)
q.push(temp->left); //EnQueue
if(temp->right)
q.push(temp->right); //EnQueue
}
cout<<endl;
return count;
}
tree* newnode(int data)  // creating new node
{ 
tree* node = (tree*)malloc(sizeof(tree)); 
node->data = data; 
node->left = NULL; 
node->right = NULL; 
return(node); 
} 
int main() 
{ 
//**same tree is builted as shown in example**
int c;
tree *root=newnode(2); 
root->left= newnode(7); 
root->right= newnode(5); 
root->right->right=newnode(9);
root->right->right->left=newnode(4);
root->left->left=newnode(2); 
root->left->right=newnode(6);
root->left->right->left=newnode(5);
root->left->right->right=newnode(11);
cout<<"printing the leaf nodes......"<<endl; 
c=noofleafnodes(root);
cout<<"no of leaf nodes is : "<<c<<endl;	
return 0; 
} 

Output

输出量

printing the leaf nodes......
2 5 11 4
no of leaf nodes is : 4

翻译自: https://www.includehelp.com/data-structure-tutorial/find-the-number-of-leaf-nodes-in-a-binary-tree.aspx

数据结构树二叉树计算节点