the recursion-tree method
If you trace a recursive algorithm on paper, the calls naturally fan out into a tree: the original problem at the root, its subproblems as children, their subproblems as grandchildren, and so on down to the base cases at the leaves. The recursion-tree method just draws this tree, labels each node with the work it does locally, and adds everything up.
Here is the procedure on T(n) = 2 T(n/2) + n. Write n at the root: that is the combine cost at the top level. The root has two children, each a problem of size n/2, each costing n/2 locally, so the next level's total is 2 * (n/2) = n. Each of those splits into two of size n/4 costing n/4, so that level totals 4 * (n/4) = n again. Every level sums to n. The tree has 1 + log2(n) levels (halving n down to 1), so the grand total is about n * (1 + log2 n) = Theta(n log n). The trick is to find the per-level sum and the number of levels, then multiply (or sum a series when the per-level totals are not constant).
The recursion tree is the best tool for forming a guess, especially when the per-level work changes. If each level's total grows geometrically (gets bigger toward the leaves), the leaves dominate and the total is Theta(leaves); if it shrinks geometrically toward the leaves, the root dominates and the total is Theta(f(n)); if it stays flat, you get a logarithmic factor. Its honest limitation: a hand-drawn tree is a heuristic. Floors, ceilings, and uneven splits make it approximate, so for a rigorous result you confirm the guessed bound with the substitution method or invoke the Master Theorem.
For T(n) = T(n/2) + T(n/4) + n the levels do NOT all sum to the same thing: level totals shrink geometrically (n, then 3n/4, then 9n/16, ...), a series that sums to 4n, so the root dominates and T(n) = Theta(n). The tree exposes this at a glance.
Sum each level, then sum the levels; geometric growth means leaves win, geometric decay means the root wins.
A recursion tree gives a guess, not a proof — the leaf count and level totals are easy to miscount with floors or unequal splits. Confirm with substitution before claiming the bound is exact.