Given a height h, count and return the maximum number of balanced binary trees possible with height h. A balanced binary tree is one in which for every node, the difference between heights of left and right subtree is not more than 1.**Examples :**

Input : h = 3 Output : 15 Input : h = 4 Output : 315

Following are the balanced binary trees of height 3.

Height of tree, h = 1 + max(left height, right height)

Since the difference between the heights of left and right subtree is not more than one, possible heights of left and right part can be one of the following:

- (h-1), (h-2)
- (h-2), (h-1)
- (h-1), (h-1)

count(h) = count(h-1) * count(h-2) + count(h-2) * count(h-1) + count(h-1) * count(h-1) = 2 * count(h-1) * count(h-2) + count(h-1) * count(h-1) = count(h-1) * (2*count(h - 2) + count(h - 1))

Hence we can see that the problem has optimal substructure property.

A **recursive function** to count no of balanced binary trees of height h is:

int countBT(int h) { // One tree is possible with height 0 or 1 if (h == 0 || h == 1) return 1; return countBT(h-1) * (2 *countBT(h-2) + countBT(h-1)); }

The time complexity of this recursive approach will be exponential. The recursion tree for the problem with h = 3 looks like :

As we can see, sub-problems are solved repeatedly. Therefore we store the results as we compute them.

An efficient dynamic programming approach will be as follows :

Below is the implementation of above approach:

## C++

`// C++ program to count number of balanced` `// binary trees of height h.` `#include <bits/stdc++.h>` `#define mod 1000000007` `using` `namespace` `std;` ` ` `long` `long` `int` `countBT(` `int` `h) {` ` ` ` ` `long` `long` `int` `dp[h + 1];` ` ` `//base cases` ` ` `dp[0] = dp[1] = 1;` ` ` `for` `(` `int` `i = 2; i <= h; i++) {` ` ` `dp[i] = (dp[i - 1] * ((2 * dp [i - 2])%mod + dp[i - 1])%mod) % mod;` ` ` `}` ` ` `return` `dp[h];` `}` `// Driver program` `int` `main()` `{` ` ` `int` `h = 3;` ` ` `cout << ` `"No. of balanced binary trees"` ` ` `" of height h is: "` ` ` `<< countBT(h) << endl;` `}` |

## Java

`// Java program to count number of balanced` `// binary trees of height h.` `class` `GFG {` ` ` ` ` `static` `final` `int` `MOD = ` `1000000007` `;` ` ` ` ` `public` `static` `long` `countBT(` `int` `h) {` ` ` `long` `[] dp = ` `new` `long` `[h + ` `1` `];` ` ` ` ` `// base cases` ` ` `dp[` `0` `] = ` `1` `;` ` ` `dp[` `1` `] = ` `1` `;` ` ` ` ` `for` `(` `int` `i = ` `2` `; i <= h; ++i)` ` ` `dp[i] = (dp[i - ` `1` `] * ((` `2` `* dp [i - ` `2` `])% MOD + dp[i - ` `1` `]) % MOD) % MOD;` ` ` ` ` `return` `dp[h];` ` ` `}` ` ` ` ` `// Driver program` ` ` `public` `static` `void` `main (String[] args) {` ` ` `int` `h = ` `3` `;` ` ` `System.out.println(` `"No. of balanced binary trees of height "` `+h+` `" is: "` `+countBT(h));` ` ` `}` `}` `/*` `This code is contributed by` `Brij Raj Kishore` `*/` |

## Python3

`# Python3 program to count number of balanced` `# binary trees of height h.` `def` `countBT(h) :` ` ` `MOD ` `=` `1000000007` ` ` `#initialize list` ` ` `dp ` `=` `[` `0` `for` `i ` `in` `range` `(h ` `+` `1` `)]` ` ` ` ` `#base cases` ` ` `dp[` `0` `] ` `=` `1` ` ` `dp[` `1` `] ` `=` `1` ` ` ` ` `for` `i ` `in` `range` `(` `2` `, h ` `+` `1` `) :` ` ` `dp[i] ` `=` `(dp[i ` `-` `1` `] ` `*` `((` `2` `*` `dp [i ` `-` `2` `])` `%` `MOD ` `+` `dp[i ` `-` `1` `])` `%` `MOD) ` `%` `MOD` ` ` ` ` `return` `dp[h]` `#Driver program` `h ` `=` `3` `print` `(` `"No. of balanced binary trees of height "` `+` `str` `(h)` `+` `" is: "` `+` `str` `(countBT(h)))` `# This code is contributed by` `# Brij Raj Kishore` |

## C#

`// C# program to count number of balanced` `// binary trees of height h.` `using` `System;` `class` `GFG {` ` ` ` ` `static` `int` `MOD = 1000000007;` ` ` ` ` `public` `static` `long` `countBT(` `int` `h) {` ` ` `long` `[] dp = ` `new` `long` `[h + 1];` ` ` ` ` `// base cases` ` ` `dp[0] = 1;` ` ` `dp[1] = 1;` ` ` ` ` `for` `(` `int` `i = 2; i <= h; ++i)` ` ` `dp[i] = (dp[i - 1] * ((2 * dp [i - 2])% MOD + dp[i - 1]) % MOD) % MOD;` ` ` ` ` `return` `dp[h];` ` ` `}` ` ` ` ` `// Driver program` ` ` `static` `void` `Main () {` ` ` `int` `h = 3;` ` ` `Console.WriteLine(` `"No. of balanced binary trees of height "` `+h+` `" is: "` `+countBT(h));` ` ` `}` ` ` `// This code is contributed by Ryuga` `}` |

## PHP

`<?php` `// PHP program to count` `// number of balanced` `$mod` `=1000000007;` `function` `countBT(` `$h` `)` `{` ` ` `global` `$mod` `;` ` ` ` ` `// base cases` ` ` `$dp` `[0] = ` `$dp` `[1] = 1;` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$h` `; ` `$i` `++)` ` ` `{` ` ` `$dp` `[` `$i` `] = (` `$dp` `[` `$i` `- 1] *` ` ` `((2 * ` `$dp` `[` `$i` `- 2]) %` ` ` `$mod` `+ ` `$dp` `[` `$i` `- 1]) %` ` ` `$mod` `) % ` `$mod` `;` ` ` `}` ` ` `return` `$dp` `[` `$h` `];` `}` `// Driver Code` `$h` `= 3;` `echo` `"No. of balanced binary trees"` `,` ` ` `" of height h is: "` `,` ` ` `countBT(` `$h` `) ,` `"\n"` `;` `// This code is contributed by aj_36` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to count number of balanced binary trees of height h.` ` ` ` ` `let MOD = 1000000007;` ` ` ` ` `function` `countBT(h) {` ` ` `let dp = ` `new` `Array(h + 1);` ` ` `dp.fill(0);` ` ` ` ` `// base cases` ` ` `dp[0] = 1;` ` ` `dp[1] = 1;` ` ` ` ` `for` `(let i = 2; i <= h; ++i)` ` ` `dp[i] = (dp[i - 1] * ((2 * dp [i - 2])% MOD + dp[i - 1]) % MOD) % MOD;` ` ` ` ` `return` `dp[h];` ` ` `}` ` ` ` ` `let h = 3;` ` ` `document.write(` `"No. of balanced binary trees of height h is: "` `+countBT(h));` ` ` `</script>` |

**Output :**

No of balanced binary trees of height h is: 15

This article is contributed by **Aditi Sharma**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **DSA Live Classes**