If it is so, then we can apply binary search on the last $\left \lceil \frac n2 \right \rceil$elements (representing the leaf nodes) to get the minimum element, which results in $log\left ( \frac n2 \right )$ =$O\left ( log n \right )$ time complexity.

Edit: My bad. Just realized that the last $\left \lceil \frac n2 \right \rceil$ elements are not in sorted order, so can’t apply binary search without sorting them.