Input: 0 / \ 1 1 / \ / \ 1 10 70 1 / \ 81 2 Output: 2 Explanation: There are 2 Fibonacci path for the above Binary Tree, for x = 3, Path 1: 0 1 1 Path 2: 0 1 1 2 Input: 8 / \ 4 81 / \ / \ 3 2 70 243 / \ 81 909 Output: 0
Approach: The idea is to use Preorder Tree Traversal. During preorder traversal of the given binary tree do the following:
- First calculate the Height of binary tree .
- Now create a vector of length equals height of tree, such that it contains Fibonacci Numbers.
- If current value of the node at ith level is not equal to ith term of fibonacci series or pointer becomes NULL then return the count.
- If the current node is a leaf node then increment the count by 1.
- Recursively call for the left and right subtree with the updated count.
- After all recursive call, the value of count is number of Fibonacci paths for a given binary tree.
Below is the implementation of the above approach:
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- Count the Number of Binary Search Trees present in a Binary Tree
- Count of 1's in any path in a Binary Tree
- Count of subtrees in a Binary Tree having XOR value K
- Count of subtrees in a Binary Tree having bitwise OR value K
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