Leaf And Path

Moderate
0/80
Average time to solve is 35m
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Asked in company
Amazon

Problem statement

You are given a binary tree of 'N' nodes. Your task is to check if there exists any root to leaf path with the same representation as the leaf traversal of the tree.

Both, the root to leaf path and the leaf traversal, are obtained by concatenating the data of the nodes in the path.

For example:

If the path is 1 -> 20 -> 8 -> 721, then the sequence obtained is “1208721”.

Leaf traversal is the sequence of leaves of the tree traversed from left to right.

Detailed explanation ( Input/output format, Notes, Images )
Input Format : 
The first line of input contains an integer 'T' representing the number of test cases. Then the test cases follow :

The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. 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 :

Example Input

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).
Note : 
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
Output Format : 
For each test case, print “True” if a path with the same representation as the leaf traversal exists, else “False”.

The output of each test case will be printed in a separate line.
Note : 
You don't need to print anything. It has already been taken care of, just implement the given function.
Constraints : 
1 <= T <= 2*10^2
0 <= N <= 3*10^3
0 <= NODE.DATA <= 10^5

Time Limit: 1 sec
Sample Input 1 :
1
2 35 10 2 3 5 2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
Sample Output 1 :
True
Explanation For Sample Input 1 :

Sample Input 1

The leaf traversal for the above binary tree is “2352”. The path 2->35->2 has the same representation as the leaf traversal i.e. “2352”. Thus, the answer is true.
Sample Input 2 :
2
1 2 3 4 5 -1 7 -1 -1 -1 -1 -1 -1
5 -1 -1
Sample Output 2 :
False
True
Hint

Try to solve the problem by first finding the leaf traversal and then checking if there exists a path with the same representation.

Approaches (1)
Brute Force

The main idea behind this approach is to solve the problem in 2 parts:

 

  1. Finding the leaf traversal for the given tree and the recursive algorithm for finding the leaf traversal is as follows :
    • Check if the current node is ‘NULL’. If yes, then return.
    • Check if it is a leaf node. If yes, then add it to the leaf traversal (the leaf traversal is stored in a string say, ‘valString’)
    • In the above step, if the node is not a leaf, then call the same function recursively for the left and the right child of the node.
       
  2. Checking if a path with the same representation as the leaf traversal exists :
    • We use recursion for this part also and start traversing the tree in preorder fashion.
    • After adding the value of the current node to the path (which is stored in a string, ‘valString’), we check if it is following the requirement (i.e. the current path is a prefix of the leaf traversal) or not.
    • If it is, then it means that the current path may be identical to the leaf traversal and we move further down the tree.
    • If not, then it means that the current path cannot be identical to the leaf traversal and we return false.
    • If we reach a situation where by adding the current node to the path gives us the sequence identical to the leaf traversal, we check if the current node is a leaf node.
    • If it is a leaf node, then the answer is “True”.
    • Else, if even after the traversal of the entire tree no such path is found, the answer is “False”.
Time Complexity

O(N) , where ‘N’ is the number of nodes in the tree.

 

We traverse the entire tree once, both for finding the leaf traversal and checking if a path with the same representation as the leaf traversal exists. Hence, the overall time complexity will be O(N+N) i.e. O(N).

Space Complexity

O(N), where ‘N’ is the number of nodes in the tree.

 

The recursive stack will take O(H) space where ‘H’ is the height of the tree. In the worst case, ‘H’ will equal to ‘N’, when given tree is skewed tree. Hence, the overall space complexity will be O(N).

Code Solution
(100% EXP penalty)
Leaf And Path
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