Euler tour of tree has been already discussed which flattens the hierarchical structure of tree into array which contains exactly 2*N-1 values. In this post, the task is to prove that the degree of Euler Tour Tree is 2 times the number of nodes minus one. Here degree means the total number of nodes get traversed in Euler Tour.
Using Example 1:
It can be seen that each node’s count in Euler Tour is exactly equal to the out-degree of node plus 1.
Mathematically, it can be represented as:
Total represents total number of nodes in Euler Tour Tree
represents ith node in given Tree
N represents the total number of node in given Tree
represents number of child of
Calculated Answer is 15 and is Equal to Actual Answer Calculated Answer is 17 and is Equal to Actual Answer
- Euler Tour of Tree
- Euler tour of Binary Tree
- Euler Tour | Subtree Sum using Segment Tree
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Total sum except adjacent of a given node in a Binary Tree
- XOR of all the nodes in the sub-tree of the given node
- Sum of all nodes in a binary tree
- Subtree of all nodes in a tree using DFS
- XOR of path between any two nodes in a Binary Tree
- Number of nodes greater than a given value in n-ary tree
- Count greater nodes in AVL tree
- Sink even nodes in Binary Tree
- Print all leaf nodes of an n-ary tree using DFS
- Sum of all the Boundary Nodes of a Binary Tree
- Count the nodes in the given tree whose weight is even
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