We are given a tree of size n as array [0..n-1] where every index i in [] represents a node and the value at i represents the immediate of that node. For root node value will be -1. Find the of the  tree given the parent links.

Examples:

```Input : parent[] = {-1, 0, 0, 0, 3, 1, 1, 2}
Output : 2

Input  : parent[] = {-1, 0, 1, 2, 3}
Output : 4
```

Here,  generic tree is sometimes also called as  N-ary tree or N-way tree where N denotes the maximum number of child a node can have. In this problem array represents n number of nodes in the tree.

Approach 1:
One solution is to traverse up the tree from node till root node is reached with node value -1. While Traversing for each node store maximum path length.
Complexity of this solution is O(n^2).

Approach 2:
Build graph for N-ary Tree in O(n) time and apply BFS on the stored graph in O(n) time and while doing BFS store maximum reached level. This solution does two iterations to find the height of N-ary tree.

```// C++ code to find height of N-ary
// tree in O(n)
#include <bits/stdc++.h>
#define MAX 1001
using namespace std;

// store N-ary tree

// Build tree in tree in O(n)
int build_tree(int arr[], int n)
{
int root_index = 0;

// Iterate for all nodes
for (int i = 0; i < n; i++) {

// if root node, store index
if (arr[i] == -1)
root_index = i;

else {
}
}
return root_index;
}

// Applying BFS
int BFS(int start)
{
// map is used as visited array
map<int, int> vis;

queue<pair<int, int> > q;
int max_level_reached = 0;

// height of root node is zero
q.push({ start, 0 });

// p.first denotes node in adjacency list
// p.second denotes level of p.first
pair<int, int> p;

while (!q.empty()) {

p = q.front();
vis[p.first] = 1;

// store the maximum level reached
max_level_reached = max(max_level_reached,
p.second);

q.pop();

for (int i = 0; i < adj[p.first].size(); i++)

// adding 1 to previous level
// stored on node p.first
// which is parent of node adj[p.first][i]
// if adj[p.first][i] is not visited
q.push({ adj[p.first][i], p.second + 1 });
}

return max_level_reached;
}

// Driver Function
int main()
{
// node 0 to node n-1
int parent[] = { -1, 0, 1, 2, 3 };

// Number of nodes in tree
int n = sizeof(parent) / sizeof(parent[0]);

int root_index = build_tree(parent, n);

int ma = BFS(root_index);
cout << "Height of N-ary Tree=" << ma;
return 0;
}
```
Output:
```Height of N-ary Tree=4
```

Time Complexity of this solution is O(2n) which converges to O(n) for very large n.

Approach 3:
We can find the height of N-ary Tree in only one iteration. We visit nodes from 0 to n-1 iteratively and mark the unvisited ancestors recursively if they are not visited before till we reach a node which is visited or we reach root node. If we reach visited node while traversing up the tree using parent links, then we use its height and will not go further in recursion.

Explanation For Example 1::

For node 0 : Check for Root node is true,
Return 0 as height, Mark node 0 as visited
For node 1 : Recur for immediate ancestor, i.e 0, which is already visited
So, Use it’s height and return height(node 0) +1
Mark node 1 as visited
For node 2 : Recur for immediate ancestor, i.e 0, which is already visited
So, Use it’s height and return height(node 0) +1
Mark node 2 as visited
For node 3 : Recur for immediate ancestor, i.e 0, which is already visited
So, Use it’s height and return height(node 0) +1
Mark node 3 as visited
For node 4 : Recur for immediate ancestor, i.e 3, which is already visited
So, Use it’s height and return height(node 3) +1
Mark node 3 as visited
For node : Recur for immediate ancestor, i.e 1, which is already visited
So, Use it’s height and return height(node 1) +1
Mark node 5 as visited
For node 6 : Recur for immediate ancestor, i.e 1, which is already visited
So, Use it’s height and return height(node 1) +1
Mark node 6 as visited
For node 7 : Recur for immediate ancestor, i.e 2, which is already visited
So, Use it’s height and return height(node 2) +1
Mark node 7 as visited

Hence, we processed each node in N-ary tree only once.

```// C++ code to find height of N-ary
// tree in O(n) (Efficient Approach)
#include <bits/stdc++.h>
using namespace std;

// Recur For Ancestors of node and
// store height of node at last
int fillHeight(int p[], int node, int visited[],
int height[])
{
// If root node
if (p[node] == -1) {

// mark root node as visited
visited[node] = 1;
return 0;
}

// If node is already visited
if (visited[node])
return height[node];

// Visit node and calculate its height
visited[node] = 1;

// recur for the parent node
height[node] = 1 + fillHeight(p, p[node],
visited, height);

// return calculated height for node
return height[node];
}

int findHeight(int parent[], int n)
{
// To store max height
int ma = 0;

// To check whether or not node is visited before
int visited[n];

// For Storing Height of node
int height[n];

memset(visited, 0, sizeof(visited));
memset(height, 0, sizeof(height));

for (int i = 0; i < n; i++) {

// If not visited before
if (!visited[i])
height[i] = fillHeight(parent, i,
visited, height);

// store maximum height so far
ma = max(ma, height[i]);
}

return ma;
}

// Driver Function
int main()
{
int parent[] = { -1, 0, 0, 0, 3, 1, 1, 2 };
int n = sizeof(parent) / sizeof(parent[0]);

cout << "Height of N-ary Tree = "
<< findHeight(parent, n);
return 0;
}
```
Output:
```Height of N-ary Tree = 2
```

Time Complexity: O(n)

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