Weighted binary search tree Suppose we have binary search tree with $n$ nodes. For each node $u$
we define weight of $u$ is number of nodes in sub tree of $u$ with
$u$. if  ratio of weight of each internal node $u$  of right sub tree
and left sub tree be at least $0.5$ and at most 2, then what is worst
case of searching for a element in such binary search tree?
The answer is $2\log n$.

How it be $2\log n$? I think this tree is weighted binary search tree not AVL tree. But in proving worst case that mentioned get  stuck. Any hint be appreciated.
My attempt:
I try to use formula that we use for minimum number of nodes we need to construct a AVL tree, Prove that AVL tree has this kind of property with Fibonacci sequence, but i can't draw any relation between my problem and AVL tree.
Show by induction on the height $h$ of the tree that a tree of height $h$ has at least $2^{\frac{h}{2}}$ nodes.
The base case $h=0$ is easy since a tree of height $0$ has $1$ vertex and $1 \ge 2^\frac{0}{2}$.
Suppose that the claim holds up to some $h \ge 0$. Consider a tree $T$ with height $h+1$ rooted in $r$. Let $u$ and $v$ be the children of $r$. Let $n(x)$ and $h(x)$ denote the number of nodes and the height of the subtree of $T$ rooted at $x$, respectvely.
We must have $h(r) = 1 +\max\{h(u), h(v)\}$. Suppose w.l.o.g. that $h(r)=1+h(u)$. Then, regardless of whether $u$ is the left or right child of $r$, we have $\frac{n(v)}{n(u)} \ge \frac{1}{2}$, i.e., $n(v) \ge \frac{n(u)}{2}$.
$$
\begin{align*}
n(r) &amp;= 1 + n(u) + n(v) &gt; n(u) + \frac{n(u)}{2} = \frac{3}{2} \cdot n(u) \\
&amp;\ge \frac{3}{2} \cdot 2^{\frac{h(u)}{2}} 
= \frac{3}{2} \cdot 2^{\frac{h(r)-1}{2}}
= \frac{3}{2} \cdot 2^{\frac{h}{2}} 
&gt; 2^{\frac{1}{2}} \cdot 2^{\frac{h}{2}} =  2^{\frac{h+1}{2}}.
\end{align*}
$$
This concludes the proofs, and shows that a tree with $n$ nodes has height at most $2 \log n$.