如果我们使用循环而不是 递归或者反之，在算法中两者都可以达到相同的目的?”

Usually yes if you are writing in a imperative language iteration will run faster than recursion, the performance hit is minimized in problems where the iterative solution requires manipulating Stacks and popping items off of a stack due to the recursive nature of the problem. There are a lot of times where the recursive implementation is much easier to read because the code is much shorter, so you do want to consider maintainability. Especailly in cases where the problem has a recursive nature. So take for example:

河内塔的递归实现:

def TowerOfHanoi(n , source, destination, auxiliary):
    if n==1:
        print ("Move disk 1 from source",source,"to destination",destination)
        return
    TowerOfHanoi(n-1, source, auxiliary, destination)
    print ("Move disk",n,"from source",source,"to destination",destination)
    TowerOfHanoi(n-1, auxiliary, destination, source)

相当短，很容易读。将其与对应的迭代TowerOfHanoi进行比较:

# Python3 program for iterative Tower of Hanoi
import sys
 
# A structure to represent a stack
class Stack:
    # Constructor to set the data of
    # the newly created tree node
    def __init__(self, capacity):
        self.capacity = capacity
        self.top = -1
        self.array = [0]*capacity
 
# function to create a stack of given capacity.
def createStack(capacity):
    stack = Stack(capacity)
    return stack
  
# Stack is full when top is equal to the last index
def isFull(stack):
    return (stack.top == (stack.capacity - 1))
   
# Stack is empty when top is equal to -1
def isEmpty(stack):
    return (stack.top == -1)
   
# Function to add an item to stack.
# It increases top by 1
def push(stack, item):
    if(isFull(stack)):
        return
    stack.top+=1
    stack.array[stack.top] = item
   
# Function to remove an item from stack.
# It decreases top by 1
def Pop(stack):
    if(isEmpty(stack)):
        return -sys.maxsize
    Top = stack.top
    stack.top-=1
    return stack.array[Top]
   
# Function to implement legal
# movement between two poles
def moveDisksBetweenTwoPoles(src, dest, s, d):
    pole1TopDisk = Pop(src)
    pole2TopDisk = Pop(dest)
 
    # When pole 1 is empty
    if (pole1TopDisk == -sys.maxsize):
        push(src, pole2TopDisk)
        moveDisk(d, s, pole2TopDisk)
       
    # When pole2 pole is empty
    else if (pole2TopDisk == -sys.maxsize):
        push(dest, pole1TopDisk)
        moveDisk(s, d, pole1TopDisk)
       
    # When top disk of pole1 > top disk of pole2
    else if (pole1TopDisk > pole2TopDisk):
        push(src, pole1TopDisk)
        push(src, pole2TopDisk)
        moveDisk(d, s, pole2TopDisk)
       
    # When top disk of pole1 < top disk of pole2
    else:
        push(dest, pole2TopDisk)
        push(dest, pole1TopDisk)
        moveDisk(s, d, pole1TopDisk)
   
# Function to show the movement of disks
def moveDisk(fromPeg, toPeg, disk):
    print("Move the disk", disk, "from '", fromPeg, "' to '", toPeg, "'")
   
# Function to implement TOH puzzle
def tohIterative(num_of_disks, src, aux, dest):
    s, d, a = 'S', 'D', 'A'
   
    # If number of disks is even, then interchange
    # destination pole and auxiliary pole
    if (num_of_disks % 2 == 0):
        temp = d
        d = a
        a = temp
    total_num_of_moves = int(pow(2, num_of_disks) - 1)
   
    # Larger disks will be pushed first
    for i in range(num_of_disks, 0, -1):
        push(src, i)
   
    for i in range(1, total_num_of_moves + 1):
        if (i % 3 == 1):
            moveDisksBetweenTwoPoles(src, dest, s, d)
   
        else if (i % 3 == 2):
            moveDisksBetweenTwoPoles(src, aux, s, a)
   
        else if (i % 3 == 0):
            moveDisksBetweenTwoPoles(aux, dest, a, d)
 
# Input: number of disks
num_of_disks = 3
 
# Create three stacks of size 'num_of_disks'
# to hold the disks
src = createStack(num_of_disks)
dest = createStack(num_of_disks)
aux = createStack(num_of_disks)
 
tohIterative(num_of_disks, src, aux, dest)

Now the first one is way easier to read because suprise suprise shorter code is usually easier to understand than code that is 10 times longer. Sometimes you want to ask yourself is the extra performance gain really worth it? The amount of hours wasted debugging the code. Is the iterative TowerOfHanoi faster than the Recursive TowerOfHanoi? Probably, but not by a big margin. Would I like to program Recursive problems like TowerOfHanoi using iteration? Hell no. Next we have another recursive function the Ackermann function: Using recursion:

    if m == 0:
        # BASE CASE
        return n + 1
    elif m > 0 and n == 0:
        # RECURSIVE CASE
        return ackermann(m - 1, 1)
    elif m > 0 and n > 0:
        # RECURSIVE CASE
        return ackermann(m - 1, ackermann(m, n - 1))

使用迭代:

callStack = [{'m': 2, 'n': 3, 'indentation': 0, 'instrPtr': 'start'}]
returnValue = None

while len(callStack) != 0:
    m = callStack[-1]['m']
    n = callStack[-1]['n']
    indentation = callStack[-1]['indentation']
    instrPtr = callStack[-1]['instrPtr']

    if instrPtr == 'start':
        print('%sackermann(%s, %s)' % (' ' * indentation, m, n))

        if m == 0:
            # BASE CASE
            returnValue = n + 1
            callStack.pop()
            continue
        elif m > 0 and n == 0:
            # RECURSIVE CASE
            callStack[-1]['instrPtr'] = 'after first recursive case'
            callStack.append({'m': m - 1, 'n': 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
            continue
        elif m > 0 and n > 0:
            # RECURSIVE CASE
            callStack[-1]['instrPtr'] = 'after second recursive case, inner call'
            callStack.append({'m': m, 'n': n - 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
            continue
    elif instrPtr == 'after first recursive case':
        returnValue = returnValue
        callStack.pop()
        continue
    elif instrPtr == 'after second recursive case, inner call':
        callStack[-1]['innerCallResult'] = returnValue
        callStack[-1]['instrPtr'] = 'after second recursive case, outer call'
        callStack.append({'m': m - 1, 'n': returnValue, 'indentation': indentation + 1, 'instrPtr': 'start'})
        continue
    elif instrPtr == 'after second recursive case, outer call':
        returnValue = returnValue
        callStack.pop()
        continue
print(returnValue)

再说一次，递归实现更容易理解。所以我的结论是，如果问题本质上是递归的，需要操作堆栈中的项，就使用递归。

递归的内存开销更大，因为每次递归调用通常都需要将一个内存地址推入堆栈，以便稍后程序可以返回到那个地址。

尽管如此，在许多情况下，递归比循环更自然、更可读——比如在处理树的时候。在这些情况下，我建议坚持使用递归。

我认为在(非尾)递归中，每当函数被调用时，分配一个新的堆栈等都会受到性能影响(当然取决于语言)。

递归有一个缺点，使用递归编写的算法的空间复杂度为O(n)。 而迭代方法的空间复杂度为O(1)。这是使用迭代而不是递归的优点。 那我们为什么要用递归呢?

见下文。

有时使用递归编写算法更容易，而使用迭代编写相同的算法略难。在这种情况下，如果您选择遵循迭代方法，您将不得不自己处理堆栈。

你必须记住，使用太深的递归，你会遇到堆栈溢出，这取决于允许的堆栈大小。为了防止这种情况，请确保提供一些基本情况，以结束递归。

把它写成递归，或者作为练习，可能会很有趣。

但是，如果要在生产中使用该代码，则需要考虑堆栈溢出的可能性。

尾递归优化可以消除堆栈溢出，但是您是否想要经历这样的麻烦，并且您需要知道您可以指望它在您的环境中进行优化。

每次算法递归，数据大小或n减少了多少?

If you are reducing the size of data or n by half every time you recurse, then in general you don't need to worry about stack overflow. Say, if it needs to be 4,000 level deep or 10,000 level deep for the program to stack overflow, then your data size need to be roughly 24000 for your program to stack overflow. To put that into perspective, a biggest storage device recently can hold 261 bytes, and if you have 261 of such devices, you are only dealing with 2122 data size. If you are looking at all the atoms in the universe, it is estimated that it may be less than 284. If you need to deal with all the data in the universe and their states for every millisecond since the birth of the universe estimated to be 14 billion years ago, it may only be 2153. So if your program can handle 24000 units of data or n, you can handle all data in the universe and the program will not stack overflow. If you don't need to deal with numbers that are as big as 24000 (a 4000-bit integer), then in general you don't need to worry about stack overflow.

但是，如果每次递归时都将数据或n的大小减小一个常数，那么当n仅变为20000时，就会遇到堆栈溢出。也就是说，当n为1000时，程序运行良好，你认为程序很好，然后在未来的某个时候，当n为5000或20000时，程序堆栈溢出。

所以如果你有堆栈溢出的可能，试着让它成为一个迭代的解决方案。