Javascript与python:具有两个递归分支的函数的不同输出

2024-09-28 22:34:26 发布

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我一直在将python代码翻译成javascript,没有任何问题,但是对于下面的示例,我没有这样做,不知道原因是什么。尽管有类似的代码,javascript代码产生的输出似乎与python完全不同。我已经调查了这个问题,javascript似乎没有执行第二个递归分支?有人知道如何使javascript代码输出与python类似吗?谢谢

1-Python代码:

def createNode(_id, right, left):
    return {'id': _id, 'left': left, 'right': right}

n1 = createNode(1, None, None)
n2 = createNode(2, None, None)
n3 = createNode(3, None, None)
n4 = createNode(4, None, None)
n5 = createNode(1, n1, n2)
n6 = createNode(2, n3, n4)
n7 = createNode(3, n5, n6)

def DownTree(lst, node): 
    left = node['left']
    right = node['right'] 

    if(right == None):
        lst.append(node['id'])
    else:
        DownTree(lst, right)    
    if(left == None):
        lst.append(node['id'])
    else:
        print('will excute')
        DownTree(lst, left)


lst = []
DownTree(lst, n7)

print(lst)
## outputs: [1, 1, 2, 2, 3, 3, 4, 4]

2-Javascript代码:

function createNode(_id, right, left){
    return {'id': _id, 'left': left, 'right': right}
}
n1 = createNode(1, undefined, undefined)
n2 = createNode(2, undefined, undefined)
n3 = createNode(3, undefined, undefined)
n4 = createNode(4, undefined, undefined)
n5 = createNode(1, n1, n2)
n6 = createNode(2, n3, n4)
n7 = createNode(3, n5, n6)

function DownTree(lst, node){ 
  left = node['left']
  right = node['right'] 

  if(right == undefined){
    lst.push(node['id'])
  }
  else{
    DownTree(lst, right)
  }  
  if(left == undefined){
    lst.push(node['id'])
  }
  else{
    console.log('not executed!')
    DownTree(lst, left)
  }
} 

lst = []
DownTree(lst, n7)
console.log(lst)
// outputs: [1, 1, 1, 3]

Tags: 代码rightnoneidnodejavascriptleftlst