First test case:-

Rooted at 0

If the tree is rooted at vertex \(0\), vertex \(1, 3, 4\) are the leaf nodes, and the potential of vertex \(1, 3, 4\) are \(1, 1, 0\) respectively. Thus, the potential of vertex \(2\) is \(1\), and further, the potential of vertex \(0\) is \(2\).
If the tree is rooted at vertex \(1\), vertices \(3\) and \(4\) are leaves, the potential of vertex \(3\) is \(1\), and vertex \(4\) is \(0\). The potential of vertex \(2\) is \(1\), potential of vertex \(0\) is \(1\). Thus, the potential of vertex \(1\) is \(1\)
If the tree is rooted at vertex \(2\), vertex \(1, 3, 4\) are the leaves, and the potential of vertex \(1, 3, 4\) are \(1, 1, 0\) respectively. Thus, the potential of vertex \(0\) is \(1\), and further, the potential of vertex \(2\) is \(2\).
If the tree is rooted at vertex \(3\), vertices \(4\) and \(1\) are leaves, the potential of vertex \(4\) is \(0\), and vertex \(1\) is \(1\). The potential of vertex \(0\) is \(1\), potential of vertex \(2\) is \(1\). Thus, the potential of vertex \(3\) is \(1\).
If the tree is rooted at vertex \(4\), vertices \(1\) and \(3\) are leaves, the potential of vertex \(1\) is \(1\), and vertex \(3\) is \(1\). The potential of vertex \(0\) is \(1\), potential of vertex \(2\) is \(2\). Thus, the potential of vertex \(4\) is \(1\).
Thus, the answer is \([2,1,2,1,1]\).

Second test case:-

If the tree is rooted at vertex \(0\), vertex \(1\) is the only leaf, and the potential of vertex \(1\) is \(1\). Thus, the potential of vertex \(2\) is \(1\), and further, the potential of vertex \(0\) is \(1\).
If the tree is rooted at vertex \(1\), vertex \(0\) is the only leaf, and the potential of vertex \(0\) is \(0\). Thus, the potential of vertex \(2\) is \(0\), and further, the potential of vertex \(1\) is \(0\).
If the tree is rooted at vertex \(2\), vertices \(1\) and \(0\) are leaves, the potential of vertex \(1\) is \(1\), and vertex \(0\) is \(0\). Thus, the potential of vertex \(2\) is \(1\).
Thus, the answer is \([1,0,1]\).