Tree: Level Order Traversal
Problem Statement :
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space.
For example:
1
\
2
\
5
/ \
3 6
\
4
For the above tree, the level order traversal is 1 -> 2 -> 5 -> 3 -> 6 -> 4.
Input Format
You are given a function,
void levelOrder(Node * root) {
}
Constraints
1 < = Nodes in the tree <= 500
Output Format
Print the values in a single line separated by a space.
Solution :
Solution in C :
In C ++ :
/*
struct node
{
int data;
node* left;
node* right;
}*/
int height(node * root)
{
if(!root)
return 0;
int lleft=height(root->left);
int rright=height(root->right);
if(lleft > rright)
return(lleft+1);
else
return (rright+1);
}
void level(node *root,int h){
if(!root)
return ;
if(h==1)
cout<<root->data<<" ";
level(root->left,h-1);
level(root->right,h-1);
}
void LevelOrder(node * root)
{
int n=height(root);
for(int i=1;i<=n;i++){
level(root,i);
}
}
In Java :
/*
class Node
int data;
Node left;
Node right;
*/
void LevelOrder(Node root)
{
Node last = root;
Queue<Node> q1 = new LinkedList<Node>();
Queue<Node> q2 = new LinkedList<Node>();
q1.add(root);
Node cur = null;
while (!q1.isEmpty() || !q2.isEmpty()) {
while (!q1.isEmpty()) {
cur = q1.poll();
if (cur.left != null) {
q2.add(cur.left);
}
if (cur.right != null) {
q2.add(cur.right);
}
System.out.print(cur.data + " ");
}
while (!q2.isEmpty()) {
cur = q2.poll();
if (cur.left != null) {
q1.add(cur.left);
}
if (cur.right != null) {
q1.add(cur.right);
}
System.out.print(cur.data + " ");
}
}
}
In python3 :
"""
Node is defined as
self.left (the left child of the node)
self.right (the right child of the node)
self.info (the value of the node)
"""
def levelOrder(root):
myQ = [root]
while(len(myQ) > 0):
iter = myQ.pop(0)
print(iter.info,end = " ")
if (iter.left != None):
myQ.append(iter.left)
if (iter.right != None):
myQ.append(iter.right)
In C :
/* you only have to complete the function given below.
node is defined as
struct node {
int data;
struct node *left;
struct node *right;
};
*/int height(struct node* root)
{
if(!root)
return 0;
else
{
int rl=height(root->left);
int rr=height(root->right);
if(rl>rr)
return rl+1;
else
return rr+1;
}
}void printlevelorder(struct node* root,int level)
{
if(!root)
return;
else if(level==1)
printf("%d ",root->data);
else if(level>1)
{
printlevelorder(root->left,level-1);
printlevelorder(root->right,level-1);
}
}
void levelOrder( struct node *root) {
int h=height(root);
int i;
for(i=0;i<=h;i++)
{
printlevelorder(root,i);
}
}
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