# Height of a generic tree from parent array

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

**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.

Time 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; ` ` ` `// Adjacency list to ` `// store N-ary tree ` `vector<` `int` `> adj[MAX]; ` ` ` `// 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` `{ ` ` ` `adj[i].push_back(arr[i]); ` ` ` `adj[arr[i]].push_back(i); ` ` ` `} ` ` ` `} ` ` ` `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 ` ` ` `if` `(!vis[adj[p.first][i]]) ` ` ` `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; ` `} ` |

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**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 5** : 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++

`// 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; ` `} ` |

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## Python3

# Python3 code to find height of N-ary

# tree in O(n) (Efficient Approach)

# Recur For Ancestors of node and

# store height of node at last

def fillHeight(p, node, visited, 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]

def findHeight(parent, n):

# To store max height

ma = 0

# To check whether or not node is

# visited before

visited = [0] * n

# For Storing Height of node

height = [0] * n

for i in range(n):

# If not visited before

if (not visited[i]):

height[i] = fillHeight(parent, i,

visited, height)

# store maximum height so far

ma = max(ma, height[i])

return ma

# Driver Code

if __name__ == ‘__main__’:

parent = [-1, 0, 0, 0, 3, 1, 1, 2]

n = len(parent)

print(“Height of N-ary Tree =”,

findHeight(parent, n))

# This code is contributed by PranchalK

**Output:**

Height of N-ary Tree = 2

**Time Complexity:** O(n)

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