证明:
设:节点总个数为n,叶子节点个数为n0n_0n0,度为1的节点个数为n1n_1n1,度为2的节点个数为n2n_2n2,边的个数为b。
n = n0n_0n0 + n1n_1n1 + n2n_2n2
b = n - 1
可得
b = n0n_0n0 + n1n_1n1 + n2n_2n2 - 1
另
b = n1n_1n1 + 2 * n2n_2n2
则
n0n_0n0 + n1n_1n1 + n2n_2n2 - 1 = n1n_1n1 + 2 * n2n_2n2
即
n0n_0n0 = n2n_2n2 + 1
为什么二叉树中叶子节点个数等于度为2的节点个数+1
最新推荐文章于 2025-06-30 13:40:45 发布