For a given Binary Tree of integers, replace each of its data with the depth of the tree.
Root is at depth 0, hence the root data is updated with 0. Replicate the same further going down the tree in the depth of the given tree.
The first line contains an Integer 't' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of input contains the elements of the tree in the level order form separated by a single space.
If any node does not have left or right child, take -1 in its place. Refer to the example below.
Example:
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 on its place.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
For each test case, print the inorder traversal of the tree where each node is replaced by the depth of the current node.
Output for every test case will be denoted in a separate line.
You do not need to print anything, it has already been taken care of. Just implement the given function.
1 <= T <= 100
1 <= N <= 1000
-10^9 <= data <= 10^9
Time limit: 1 sec
1
2 35 10 2 3 5 2 -1 -1 -1 -1 -1 -1 -1 -1
2 1 2 0 2 1 2
1
1 2 -1 3 -1 4 -1 5 -1 -1 -1
4 3 2 1 0
Do level order traversal and update each node with its respective depth in tree.
Traverse tree starting from the root. While traversing the tree, pass the depth of the node as a parameter. We can track depth by passing it as 0 for root and increment depth for the children by 1.
Pseudo Code:
fHelp(Node root, level=0)
f(Node root)
fHelp(root, 0)
O(N), where N is the total number of nodes in the tree.
Since every node of the tree is visited once.
O(N), where N is the total number of nodes in the tree.
Since here, the recursion stack goes to a maximum depth of the tree and if it is a bamboo tree then depth of the tree is equal to the number of nodes.
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Interview problems
c++ easy solution ,beats 100% of solution
BinaryTreeNode<int> *changeval(BinaryTreeNode<int>*root,int d){
if(root==nullptr){
return nullptr;
}
root->data=d;
++d;
if (root->left) {
BinaryTreeNode<int> *t1 = changeval(root->left, d);
}
if (root->right) {
BinaryTreeNode<int> *t2 = changeval(root->right, d);
}
return root;
}
BinaryTreeNode<int> *changeToDepthTree(BinaryTreeNode<int> *root)
{
if(root==nullptr){
return nullptr;
}
root=changeval(root,0);
return root;
}
Python easy solution.Just preorder traversal
def changeToDepthTree(root,level=0):
if root==None:
return
root.data=level
changeToDepthTree(root.left,level+1)
changeToDepthTree(root.right,level+1)
return rootInterview problems
Very easy solution
void solve(BinaryTreeNode<int> *root,int d){
if(root == NULL){
return;
}
root->data = d;
solve(root->left,d+1);
solve(root->right,d+1);
}
BinaryTreeNode<int> *changeToDepthTree(BinaryTreeNode<int> *root)
{
solve(root,0);
return root;
}Interview problems
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