在给定总和K的二叉树中找到级别

Description:

描述：

The article describes how to find the level in a binary tree with given sum K? This is an interview coding problem came in Samsung, Microsoft.

本文介绍了如何在给定总和K下在二叉树中找到级别 ？这是一个面试编码问题，来自三星，微软。

Problem statement:

问题陈述：

Given a binary tree and an integer K, output the level no of the tree which sums to K. Assume root level is level 0.

给定一棵二叉树和一个整数K ，输出该树的级数之和为K。假设根级别为0 。

Solution

解

Algorithm:

算法：

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

解决此类问题的一种流行的遍历技术是级别顺序树遍历(更多内容： 二叉树上的级别顺序遍历 )，其中我们使用了经过修改的BFS概念。

1. Set a variable level=0 & declare a variable of type tree* named temp. level is a flag for the current level & temp stores tree node to be processed.

1.设置变量level = 0并声明一个名为temp的 tree *类型的变量。级别是当前级别和临时存储要处理的树节点的标志。

2. Set cur_sum=0

2.设置cur_sum = 0

3. if(!root) // root is NULL
return -1; //no such level exists

3. if(！root) //根为NULL
返回-1; //不存在这样的级别

4. q=createQueue() //to store pointers to tree node

4. q = createQueue() //存储指向树节点的指针

5. EnQueue(q,root);

5. EnQueue(q，root);

6. EnQueue(q,NULL);

6. EnQueue(q，NULL);

Every time, we EnQueue a NULL to reflect the end of current level. Same way while processing when NULL is found, it reveals that end of current level.

每次，我们将NULL排队以反映当前级别的结束。发现NULL时进行处理的方式相同，它显示当前级别的末尾。

7. while( q is not empty)
temp=DeQueue(q);
if(temp==NULL){ //end of current level
if (cur_sum==K) // current level sum is equal to K
Return level; //return level no (root is at 0 level)
else {
Set cur_sum =0; // for next level
If(q is not empty)
// to reflect end of next level which 
// will be processed in next iteration
EnQueue(q,NULL)
Increase level //increase level count
}
}
Else{
cur_sum+=temp->data; //sum the level
//do level order traversal
If(temp->left)
EnQueue(temp->left);
If(temp->right)
EnQueue (temp->right);
}
End while loop

8. If program control comes out of while loop then surely no level has a sum K. Thus print no level has sum K

8.如果程序控制从while循环中出来，那么肯定没有水平的总和为K。因此打印无级别的总和为K

Example:

例：

Here the level sums are:
For level 0: 2
For level 1: 12
For level 2: 17
For level 3: 20
Thus if K is 12 our program will print level 1

C ++代码在给定总和为K的二叉树中找到级别 (C++ code to find the level in a binary tree with given sum K)

#include <bits/stdc++.h>
using namespace std;
// tree node is defined
class tree{
public:
int data;
tree *left;
tree *right;
};
int findLevel( tree *root,int K){
queue<tree*> q;  // using stl
tree* temp;
//counting level no & initializing cur_sum
int level=0,cur_sum=0; 
//EnQueue root
q.push(root); 
//EnQueue NULL to inticate end of 0 level
q.push(NULL);
while(!q.empty()){
//DeQueueing using STL
temp=q.front(); 
q.pop();
if(temp==NULL){
//if current level sum equals to K
if(cur_sum==K) 
return level;//return level no
//ifn't then set cur_sum to 0 for further levels
cur_sum=0; 
if(!q.empty())
q.push(NULL);
level++;
}
else{
//level order traversal
cur_sum+=temp->data; //sum thd level
if(temp->left)
q.push(temp->left); //EnQueue
if(temp->right)
q.push(temp->right); //EnQueue
}
}
return -1;
}
// creating new node
tree* newnode(int data)  
{ 
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,K;
cout<<"Tree is built like the example aforesaid"<<endl;
cout<<"input K....."<<endl;
cin>>K;
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<<"finding if any level exists......"<<endl; 
c=findLevel(root,K);
if(c>=0)
cout<<"level is found and it is level "<<c<<endl;
else
cout<<"level not found, no such tree level exists";
return 0; 
}

Output

输出量

First run:
Tree is built like the example aforesaid 
input K..... 
20 
finding if any level exists......
level is found and it is level 3 
Second run:
Tree is built like the example aforesaid 
input K..... 
25 
finding if any level exists......
level not found, no such tree level exists