For a tree, the degree of node is k at most (k > = 2), and it is proved that there are at least k leaves

If there are at most s leaves, s < K, then the tree has s 1-degree nodes, 1-k-degree nodes, and the degree of the remaining nodes is at least 2
Let n be the number of nodes, then M = n-1. According to the handshake theorem, 2m = 2n-2 = ∑ D (VI) ≥ s × 1 + K × 1 + 2 (n-s-1), s ≥ K
So there are at least k leaves

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