二叉樹查找前提,分組是有序的。python
例子,從一個數組中找到某個對象的索引。數組
def search(a, t): l = 0 r = len(a) - 1 while l <= r: mid = int((l+r)/2) print(mid) if a[mid]<t: l = mid + 1 elif a[mid]==t: return mid else: r = mid-1 return None if __name__ == '__main__': r = search([0,1,2,3,4,5,6,7,8,9], 6)