New answers tagged graph-traversal
2
Since your are talking about in-ordering, I will suppose that the tree is a binary tree (otherwise I don't know how the in-order is defined).
This result of unicity can be proven by induction of the length of arrays, but only if nodes are distincts (which I will assume for the rest of the proof). Suppose $T$ to be a tree with an in-order given by $A = (a_1, ...
0
I can show the order's visit of vertices, using the algorithm Depth first search.
#include <iostream>
#include <vector>
using namespace std;
const int maximumSize=5;
vector<int> visited(maximumSize, 0);
vector<int> graph[maximumSize];
int vertices, edges, order;
void showContentVector(vector<int> input)
{
for(int i=0; i<...
0
Both the start time and the finish time of a vertex can change when edges are examined in a different order.
Here is an example of a DFS visit starting from vertex $a$. In the figure on the left the edge $(a,b)$ is examined before the edge $(a,c)$. In the figure on the right the order is reversed.
Start times are in blue, while finish times are in red.
Top 50 recent answers are included