Left View of Binary Tree
PROBLEM :
Given a Binary Tree, print Left view of it. Left view of a Binary Tree is set of nodes visible when tree is visited from Left side.
Left view of following tree is 1 2 4 8
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
Input:
The task is to complete the method which takes one argument, root of Binary Tree. There are multiple test cases. For each test case, this method will be called individually.
Output:
The function should print nodes in left view of Binary Tree.
Constraints:
1 <=T<= 30
1 <= Number of nodes<= 100
1 <= Data of a node<= 1000
Example:
Input:
2
2
1 2 R 1 3 L
4
10 20 L 10 30 R 20 40 L 20 60 R
Output:
1 3
10 20 40
There are two test casess. First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 3 and right child of 1 is 2. Second test case represents a tree with 4 edges and 5 nodes.
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SIMPLE c++ IMPLEMENTATION :
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/* A binary tree node
struct Node
{
int data;
struct Node* left, * right;
}; */
// A wrapper over leftViewUtil()
void printleft(Node *,int ,int *) ;
void leftView(struct Node *root)
{
if(root==NULL)
return ;
int no=0 ;
printleft(root,1,&no) ;
}
void printleft(Node *root,int level,int *first)
{
if(root==NULL)
return ;
if(level>(*first))
{
cout<<root->data<<" " ;
(*first)=level ;
}
printleft(root->left,level+1,&(*first)) ;
printleft(root->right,level+1,&(*first)) ;
}
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Given a Binary Tree, print Left view of it. Left view of a Binary Tree is set of nodes visible when tree is visited from Left side.
Left view of following tree is 1 2 4 8
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
Input:
The task is to complete the method which takes one argument, root of Binary Tree. There are multiple test cases. For each test case, this method will be called individually.
Output:
The function should print nodes in left view of Binary Tree.
Constraints:
1 <=T<= 30
1 <= Number of nodes<= 100
1 <= Data of a node<= 1000
Example:
Input:
2
2
1 2 R 1 3 L
4
10 20 L 10 30 R 20 40 L 20 60 R
Output:
1 3
10 20 40
There are two test casess. First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 3 and right child of 1 is 2. Second test case represents a tree with 4 edges and 5 nodes.
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------
/* A binary tree node
struct Node
{
int data;
struct Node* left, * right;
}; */
// A wrapper over leftViewUtil()
void printleft(Node *,int ,int *) ;
void leftView(struct Node *root)
{
if(root==NULL)
return ;
int no=0 ;
printleft(root,1,&no) ;
}
void printleft(Node *root,int level,int *first)
{
if(root==NULL)
return ;
if(level>(*first))
{
cout<<root->data<<" " ;
(*first)=level ;
}
printleft(root->left,level+1,&(*first)) ;
printleft(root->right,level+1,&(*first)) ;
}
---------------------------------------------------------------------------------
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