Given a Binary Tree as shown below. The task is to count all pair of adjacent nodes whose XOR is an odd number.
Initially, root will be 0, start traversing the tree. XOR of 15 and 13 will be 2 (Even) XOR of 13 and 12 will be 1 (Odd) XOR of 13 and 14 will be 5 (Even) XOR of 15 and 18 will be 13 (Odd) XOR of 18 and 17 will be 3 (Odd) XOR of 18 and 21 will be 7 (Odd) Therefore, total adjacent pairs with odd XOR = 5
- Start traversing the tree from top to down.
- Every time perform XOR operation with the current node data and its adjacent data.
- If XOR of both node is an odd number then increment the count.
Below is the implementation of the above approach:
- Given a n-ary tree, count number of nodes which have more number of children than parents
- Count pairs with Bitwise XOR as EVEN number
- Count pairs with Bitwise XOR as ODD number
- Count the nodes whose sum with X is a Fibonacci number
- Count the number of nodes at given level in a tree using BFS.
- Maximum sum of nodes in Binary tree such that no two are adjacent
- Maximum sum of nodes in Binary tree such that no two are adjacent | Dynamic Programming
- Leftover element after performing alternate Bitwise OR and Bitwise XOR operations on adjacent pairs
- Print levels with odd number of nodes and even number of nodes
- Count pairs with Odd XOR
- Count all pairs with given XOR
- Count BST nodes that lie in a given range
- Count of pairs violating BST property
- Count the nodes in the given tree whose weight is even
- Count ordered pairs of positive numbers such that their sum is S and XOR is K
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