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);
}
}
View More Similar Problems
Tree: Level Order Traversal
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 F
View Solution →Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →Tree: Huffman Decoding
Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t
View Solution →Binary Search Tree : Lowest Common Ancestor
You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b
View Solution →Swap Nodes [Algo]
A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from
View Solution →Kitty's Calculations on a Tree
Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a
View Solution →