阿Q吧 > 一颗二叉树共有25个结点，其中5个是叶子结点，则度为1的结点数为多少？

一颗二叉树共有25个结点，其中5个是叶子结点，则度为1的结点数为多少？

2025-03-16 05:54:14
推荐回答（1个）
回答1：

二叉树有如下性质：
n0 = n2 + 1 ;即叶子节点个数等于度为2节点个数+1
所以本题，叶子节点为5个，度为2的节点为5-1=4个
度为1的节点数=总节点-度为2节点-叶子节点 = 25 -5 -4 = 16个

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