Find MEX of every subtree in given Tree

Given a Generic Tree consisting of N nodes numbered from 0 to N - 1 which is rooted at node 0 and an array val[] such that the value at each node is represented by val[i],  the task for each node is to find the value of MEX of its subtree.

The MEX value of node V is defined as the smallest missing positive number in a tree rooted at node V.

Examples:

Input: N = 6, edges = {{0, 1}, {1, 2}, {0, 3}, {3, 4}, {3, 5}}, val[] = {4, 3, 5, 1, 0, 2}
Output: [6, 0, 0, 3, 1, 0]
Explanation:
             0(4)
           /    \
      1(3)    3(1)
     /         /    \
2(5)     4(0)   5(2)  

In the subtrees of:
Node 0: All the values in range [0, 5] are present, hence the smallest non-negative value not present is 6.
Node 1: The smallest non-negative value not present in subtree of node 1 is 0.
Node 2: The smallest non-negative value not present in subtree of node 2 absent is 0.
Node 3: All the values in range [0, 2] are present, hence the smallest non-negative value not present in subtree of node 3 is 3.
Node 4: The smallest non-negative value not present in subtree of node 4 is 1.
Node 5: The smallest non-negative value not present in subtree of node 5 is 0.

Approach: The given problem can be solved using DFS Traversal on the given Tree and performing the Binary Search to find the missing minimum positive integers in each node subtree. Follow the steps below to solve the problem:

• Initialize an array, say sol[] to store the MEX for each node.
• Perform the DFS Traversal from node 0 and perform the following steps:
  • Store all the nodes during DFS for each node in the vector.
  • Merge the two vectors in the sorted order for each node in the recursive calls.
  • Apply binary search on the sorted list to find the smallest non-negative integer value which is not present in the sorted array and store the value of MEX of the current subtree in the array sol[].
• After completing the above steps, print the value stored in the array sol[] as the result.

Below is the implementation of the above approach:

6 0 0 3 1 0

Time Complexity: O(N*(N + log N))
Auxiliary Space: O(N)