在二叉树中，对于每个节点，左右子树的节点数之差最多为2。如果树的高度h > 0，则树中的最小节点数为：
(A) 2小时 – 1
(B) 2小时 – 1 + 1
(C) 2小时– 1
(D) 2小时答案：(乙)
解释：
Let there be n(h) nodes at height h.
In a perfect tree where every node has exactly
two children, except leaves, following recurrence holds.
n(h) = 2*n(h-1) + 1
In given case, the numbers of nodes are two less, therefore
n(h) = 2*n(h-1) + 1 - 2
= 2*n(h-1) - 1
Now if try all options, only option (b) satisfies above recurrence.
Let us see option (B)
n(h) = 2h - 1 + 1
So if we substitute
n(h-1) = 2h-2 + 1, we should get n(h) = 2h-1 + 1
n(h) = 2*n(h-1) - 1
= 2*(2h-2 + 1) -1
= 2h-1 + 1.
这个问题的测验