二叉树遍历python

Question

def traverse(self):
    if self.root !=None:
        print('*****Traversing*****')
        print('self.root is', self.root.data)
        print('self.root.Left is', self.root.leftchild.data)
        print('self.root.Right  is', self.root.rightchild.data)
        self.traverse_in_order(self.root)
        
def traverse_in_order(self,node):
    
    if node.leftchild!=None:  #there is leftchild
        print('node.leftchild is',node.leftchild.data )
        self.traverse_in_order(node.leftchild)       #go till end of leftnode
    print('')
    print('print node ', node.data)
    print('')
    if node.rightchild!=None:
        print('Node.RIght',node.rightchild.data)
        self.traverse_in_order(node.rightchild)
bst=BST()
bst.insert(32)
bst.insert(10)
bst.insert(1)
bst.insert(19)
bst.insert(46)

bst.traverse()

Output----

*****Traversing*****
   self.root is 32
   self.root.Left is 10
   self.root.Right  is 46
   node.leftchild is 10
   node.leftchild is 1
   -------------print node  1
   -------------print node  10
   Node.RIght 19
   print node  19
   print node  32
   Node.RIght 46
   print node  46

Can anyone help with traverse_in_order function. My doubt is from root we go till last left node and then print its data ie(1 is printed understood.)How 10 is printed again.Same with rights subtree. I hope you got what iam asking. Thanks

Answer 1

"10" is not printed again. The only prints that matter are the ones that say "print node", and those are all unique and in order. The others are just debug prints to help you follow the logic. They don't count.