A tree is Continuous tree if in each root to leaf path, absolute difference between keys of two adjacent is 1. We are given a binary tree, we need to check if tree is continuous or not.
Input : 3 / \ 2 4 / \ \ 1 3 5 Output: "Yes" // 3->2->1 every two adjacent node's absolute difference is 1 // 3->2->3 every two adjacent node's absolute difference is 1 // 3->4->5 every two adjacent node's absolute difference is 1 Input : 7 / \ 5 8 / \ \ 6 4 10 Output: "No" // 7->5->6 here absolute difference of 7 and 5 is not 1. // 7->5->4 here absolute difference of 7 and 5 is not 1. // 7->8->10 here absolute difference of 8 and 10 is not 1.
The solution requires a traversal of tree. The important things to check are to make sure that all corner cases are handled. The corner cases include, empty tree, single node tree, a node with only left child and a node with only right child.
In tree traversal, we recursively check if left and right subtree are continuous. We also check if difference between keys of current node’s key and its children keys is 1. Below is the implementation of the idea.
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Improved By : andrew1234