Segment tree - query complexity -


i having problems understanding segment tree complexity. clear if have update function has change 1 node, complexity log(n). have no idea why complexity of query(a,b), (a,b) interval needs checked, log(n). can provide me intuitive / formal proof understand this?

there 4 cases when query interval (x,y)

find(r,x,y) //r node % case 1     if r.first = x , r.last = y            return {r} % case 2     if y <= r.middle         return find(r.leftchild, x, y)  % case 3     if x >= r.middle + 1          return find(r.rightchild, x, y)  % case 4     p = find(r.leftchild, x, r.middle)     q = find(r.rightchild, r.middle + 1, y)         return p union q.

intuitively, first 3 cases reduce level of tree height 1, since tree has height log n, if first 3 cases happen, running time o(log n).

for last case, find() divide problem 2 subproblems. however, assert can happen @ once. after called find(r.leftchild, x, r.middle), querying r.leftchild interval [x, r.middle]. r.middle same r.leftchild.last. if x > r.leftchild.middle, case 1; if x <= r.leftchild, call 

find ( r.leftchild.leftchild, x, r.leftchild.middle ); find ( r.leftchild.rightchild, r.leftchild.middle + 1, , r.leftchild.last );

however, second find() returns r.leftchild.rightchild.sum , therefore takes constant time, , problem not separate 2 subproblems (strictly speaking, problem separated, though 1 subproblem takes o(1) time solve).

since same analysis holds on rightchild of r, conclude after case4 happens first time, running time t(h) (h remaining level of tree) be

t(h) <= t(h-1) + c (c constant) t(1) = c

which yields:

t(h) <= c * h = o(h) = o(log n) (since h height of tree)

hence end proof.

this first time contribute, hence if there problems, please kindly point them out , edit answer.


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