Given preorder of a binary tree, calculate its depth(or height) [starting from depth 0]. The preorder is given as a string with two possible characters.
- ‘l’ denotes the leaf
- ‘n’ denotes internal node
The given tree can be seen as a full binary tree where every node has 0 or two children. The two children of a node can ‘n’ or ‘l’ or mix of both.
Examples :
Input : nlnll Output : 2 Explanation :Input : nlnnlll Output : 3
Preorder of the binary tree is given so traverse
Also, we would be given a string of char (formed of ‘n’ and ‘l’), so there is no need to implement tree also.
The recursion function would be:
1) Base Case: return 0; when tree[i] = ‘l’ or i >= strlen(tree)
2) find_depth( tree[i++] ) //left subtree
3) find_depth( tree[i++] ) //right subtree
Where i is the index of the string tree.
C++
// C++ program to find height of full binary tree // using preorder #include <bits/stdc++.h> using namespace std; // function to return max of left subtree height // or right subtree height int findDepthRec(char tree[], int n, int& index) { if (index >= n || tree[index] == 'l') return 0; // calc height of left subtree (In preorder // left subtree is processed before right) index++; int left = findDepthRec(tree, n, index); // calc height of right subtree index++; int right = findDepthRec(tree, n, index); return max(left, right) + 1; } // Wrapper over findDepthRec() int findDepth(char tree[], int n) { int index = 0; findDepthRec(tree, n, index); } // Driver program int main() { // Your C++ Code char tree[] = "nlnnlll"; int n = strlen(tree); cout << findDepth(tree, n) << endl; return 0; } |
Java
// Java program to find height // of full binary tree using // preorder import java .io.*; class GFG { // function to return max // of left subtree height // or right subtree height static int findDepthRec(String tree, int n, int index) { if (index >= n || tree.charAt(index) == 'l') return 0; // calc height of left subtree // (In preorder left subtree // is processed before right) index++; int left = findDepthRec(tree, n, index); // calc height of // right subtree index++; int right = findDepthRec(tree, n, index); return Math.max(left, right) + 1; } // Wrapper over findDepthRec() static int findDepth(String tree, int n) { int index = 0; return (findDepthRec(tree, n, index)); } // Driver Code static public void main(String[] args) { String tree = "nlnnlll"; int n = tree.length(); System.out.println(findDepth(tree, n)); } } // This code is contributed // by anuj_67. |
Python3
#Python program to find height of full binary tree # using preorder # function to return max of left subtree height # or right subtree height def findDepthRec(tree, n, index) : if (index[0] >= n or tree[index[0]] == 'l'): return 0 # calc height of left subtree (In preorder # left subtree is processed before right) index[0] += 1 left = findDepthRec(tree, n, index) # calc height of right subtree index[0] += 1 right = findDepthRec(tree, n, index) return (max(left, right) + 1) # Wrapper over findDepthRec() def findDepth(tree, n) : index = [0] return findDepthRec(tree, n, index) # Driver program to test above functions if __name__ == '__main__': tree= "nlnnlll" n = len(tree) print(findDepth(tree, n)) # This code is contributed by SHUBHAMSINGH10 |
C#
// C# program to find height of // full binary tree using preorder using System; class GFG { // function to return max of left subtree // height or right subtree height static int findDepthRec(char[] tree, int n, int index) { if (index >= n || tree[index] == 'l') return 0; // calc height of left subtree (In preorder // left subtree is processed before right) index++; int left = findDepthRec(tree, n, index); // calc height of right subtree index++; int right = findDepthRec(tree, n, index); return Math.Max(left, right) + 1; } // Wrapper over findDepthRec() static int findDepth(char[] tree, int n) { int index = 0; return (findDepthRec(tree, n, index)); } // Driver program static public void Main() { char[] tree = "nlnnlll".ToCharArray(); int n = tree.Length; Console.WriteLine(findDepth(tree, n)); } } // This code is contributed by vt_m. |
Output:
3
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