Help me understand Inorder Traversal without using

问题:

I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion.

This is what I've tried so far:

def traverseInorder(node):
    lifo = Lifo()
    lifo.push(node)
    while True:
        if node is None:
            break
        if node.left is not None:
            lifo.push(node.left)
            node = node.left
            continue
        prev = node
        while True:
            if node is None:
                break
            print node.value
            prev = node
            node = lifo.pop()
        node = prev
        if node.right is not None:
            lifo.push(node.right)
            node = node.right
        else:
            break

The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong?

I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too.

Thanks for your time!

P.S.: Definitions of Lifo and Node:

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

class Lifo:
    def __init__(self):
        self.lifo = ()
    def push(self, data):
        self.lifo = (data, self.lifo)
    def pop(self):
        if len(self.lifo) == 0:
            return None
        ret, self.lifo = self.lifo
        return ret

回答1:

Start with the recursive algorithm (pseudocode) :

traverse(node):
  if node != None do:
    traverse(node.left)
    print node.value
    traverse(node.right)
  endif

This is a clear case of tail recursion, so you can easily turn it into a while-loop.

traverse(node):
  while node != None do:
    traverse(node.left)
    print node.value
    node = node.right
  endwhile

You're left with a recursive call. What the recursive call does is push a new context on the stack, run the code from the beginning, then retrieve the context and keep doing what it was doing. So, you create a stack for storage, and a loop that determines, on every iteration, whether we're in a "first run" situation (non-null node) or a "returning" situation (null node, non-empty stack) and runs the appropriate code:

traverse(node):
  stack = []
  while !empty(stack) || node != None do:
    if node != None do: // this is a normal call, recurse
      push(stack,node)
      node = node.left
    else // we are now returning: pop and print the current node
      node = pop(stack)
      print node.value
      node = node.right
    endif
  endwhile

The hard thing to grasp is the "return" part: you have to determine, in your loop, whether the code you're running is in the "entering the function" situation or in the "returning from a call" situation, and you will have an if/else chain with as many cases as you have non-terminal recursions in your code.

In this specific situation, we're using the node to keep information about the situation. Another way would be to store that in the stack itself (just like a computer does for recursion). With that technique, the code is less optimal, but easier to follow

traverse(node):
  // entry:
  if node == NULL do return
  traverse(node.left)
  // after-left-traversal:
  print node.value
  traverse(node.right)

traverse(node):
   stack = [node,'entry']
   while !empty(stack) do:
     [node,state] = pop(stack)
     switch state: 
       case 'entry': 
         if node == None do: break; // return
         push(stack,[node,'after-left-traversal']) // store return address
         push(stack,[node.left,'entry']) // recursive call
         break;
       case 'after-left-traversal': 
         print node.value;
         // tail call : no return address
         push(stack,[node.right,'entry']) // recursive call
      end
    endwhile 

回答2:

Here is a simple in-order non-recursive c++ code ..

void inorder (node *n)
{
    stack s;

    while(n){
        s.push(n);
        n=n->left;
    }

    while(!s.empty()){
        node *t=s.pop();
        cout<<t->data;
        t=t->right;

        while(t){
            s.push(t);
            t = t->left;
        }
    }
}

回答3:

def print_tree_in(root):
    stack = []
    current = root
    while True:
        while current is not None:
            stack.append(current)
            current = current.getLeft();
        if not stack:
            return
        current = stack.pop()
        print current.getValue()
        while current.getRight is None and stack:
            current = stack.pop()
            print current.getValue()
        current = current.getRight();

回答4:

def traverseInorder(node):
   lifo = Lifo()

  while node is not None:
    if node.left is not None:
       lifo.push(node)
       node = node.left
       continue

   print node.value

   if node.right is not None:
      node = node.right
      continue

   node = lifo.Pop()
   if node is not None :
      print node.value
      node = node.right

PS: I don't know Python so there may be a few syntax issues.

回答5:

Here is a sample of in order traversal using stack in c# (.net):

(for post order iterative you may refer to: Post order traversal of binary tree without recursion)

public string InOrderIterative()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
                var iterativeNode = this._root;
                while(iterativeNode != null)
                {
                    stack.Push(iterativeNode);
                    iterativeNode = iterativeNode.Left;
                }
                while(stack.Count > 0)
                {
                    iterativeNode = stack.Pop();
                    nodes.Add(iterativeNode.Element);
                    if(iterativeNode.Right != null)
                    {
                        stack.Push(iterativeNode.Right);
                        iterativeNode = iterativeNode.Right.Left;
                        while(iterativeNode != null)
                        {
                            stack.Push(iterativeNode);
                            iterativeNode = iterativeNode.Left;
                        }
                    }
                }
            }
            return this.ListToString(nodes);
        }

Here is a sample with visited flag:

public string InorderIterative_VisitedFlag()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
                BinaryTreeNode iterativeNode = null;
                stack.Push(this._root);
                while(stack.Count > 0)
                {
                    iterativeNode = stack.Pop();
                    if(iterativeNode.visted)
                    {
                        iterativeNode.visted = false;
                        nodes.Add(iterativeNode.Element);
                    }
                    else
                    {
                        iterativeNode.visted = true;
                        if(iterativeNode.Right != null)
                        {
                            stack.Push(iterativeNode.Right);
                        }
                        stack.Push(iterativeNode);
                        if (iterativeNode.Left != null)
                        {
                            stack.Push(iterativeNode.Left);
                        }
                    }
                }
            }
            return this.ListToString(nodes);
        }

the definitions of the binarytreenode, listtostring utility:

string ListToString(List<int> list)
        {
            string s = string.Join(", ", list);
            return s;
        }


class BinaryTreeNode
    {
        public int Element;
        public BinaryTreeNode Left;
        public BinaryTreeNode Right;        
    }

回答6:

State can be remembered implicitly,

traverse(node) {
   if(!node) return;
   push(stack, node);
   while (!empty(stack)) {
     /*Remember the left nodes in stack*/
     while (node->left) {
        push(stack, node->left);
        node = node->left;
      }

      /*Process the node*/
      printf("%d", node->data);

      /*Do the tail recursion*/
      if(node->right) {
         node = node->right
      } else {
         node = pop(stack); /*New Node will be from previous*/
      }
    }
 }

回答7:

@Victor, I have some suggestion on your implementation trying to push the state into the stack. I don't see it is necessary. Because every element you take from the stack is already left traversed. so instead of store the information into the stack, all we need is a flag to indicate if the next node to be processed is from that stack or not. Following is my implementation which works fine:

def intraverse(node):
    stack = []
    leftChecked = False
    while node != None:
        if not leftChecked and node.left != None:
            stack.append(node)
            node = node.left
        else:
            print node.data
            if node.right != None:
                node = node.right
                leftChecked = False
            elif len(stack)>0:
                node = stack.pop()
                leftChecked = True
            else:
                node = None

回答8:

Little Optimization of answer by @Emadpres

def in_order_search(node):
    stack = Stack()
    current = node

    while True:
        while current is not None:
            stack.push(current)
            current = current.l_child

        if stack.size() == 0:
            break

        current = stack.pop()
        print(current.data)
        current = current.r_child

回答9:

This may be helpful (Java implementation)

public void inorderDisplay(Node root) {
    Node current = root;
    LinkedList<Node> stack = new LinkedList<>();
    while (true) {
        if (current != null) {
            stack.push(current);
            current = current.left;
        } else if (!stack.isEmpty()) {
            current = stack.poll();
            System.out.print(current.data + " ");
            current = current.right;
        } else {
            break;
        }
    }
}

回答10:

Simple iterative inorder traversal without recursion

'''iterative inorder traversal, O(n) time & O(n) space '''

class Node:
    def __init__(self, value, left = None, right = None):
        self.value = value
        self.left = left
        self.right = right

def inorder_iter(root):

    stack = [root]
    current = root

    while len(stack) > 0:
        if current:
            while current.left:
                stack.append(current.left)
                current = current.left
        popped_node = stack.pop()
        current = None
        if popped_node:
            print popped_node.value
            current = popped_node.right
            stack.append(current)

a = Node('a')
b = Node('b')
c = Node('c')
d = Node('d')

b.right = d
a.left = b
a.right = c

inorder_iter(a)

回答11:

class Tree:

    def __init__(self, value):
        self.left = None
        self.right = None
        self.value = value

    def insert(self,root,node):
        if root is None:
            root = node
        else:
            if root.value < node.value:
                if root.right is None:
                    root.right = node
                else:
                    self.insert(root.right, node)
            else:
                if root.left is None:
                    root.left = node
                else:
                    self.insert(root.left, node)       

    def inorder(self,tree):
        if tree.left != None:
            self.inorder(tree.left)
        print "value:",tree.value

        if tree.right !=None:
            self.inorder(tree.right)

    def inorderwithoutRecursion(self,tree):
        holdRoot=tree
        temp=holdRoot
        stack=[]
        while temp!=None:
            if temp.left!=None:
                stack.append(temp)
                temp=temp.left
                print "node:left",temp.value

            else:
                if len(stack)>0:
                    temp=stack.pop();
                    temp=temp.right
                    print "node:right",temp.value

回答12:

Here's an iterative C++ solution as an alternative to what @Emadpres posted:

void inOrderTraversal(Node *n)
{
    stack<Node *> s;
    s.push(n);
    while (!s.empty()) {
        if (n) {
            n = n->left;
        } else {
            n = s.top(); s.pop();
            cout << n->data << " ";
            n = n->right;
        }
        if (n) s.push(n);
    }
}

回答13:

Here is an iterative Python Code for Inorder Traversal ::

class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

def inOrder(root):
    current = root
    s = []
    done = 0

    while(not done):
        if current is not None :
            s.append(current)
            current = current.left
        else :
            if (len(s)>0):
                current = s.pop()
                print current.data
                current = current.right
            else :
                done =1

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)

inOrder(root)

回答14:

I think part of the problem is the use of the "prev" variable. You shouldn't have to store the previous node you should be able to maintain the state on the stack (Lifo) itself.

From Wikipedia, the algorithm you are aiming for is:

1. Visit the root.
2. Traverse the left subtree
3. Traverse the right subtree

In pseudo code (disclaimer, I don't know Python so apologies for the Python/C++ style code below!) your algorithm would be something like:

lifo = Lifo();
lifo.push(rootNode);

while(!lifo.empty())
{
    node = lifo.pop();
    if(node is not None)
    {
        print node.value;
        if(node.right is not None)
        {
            lifo.push(node.right);
        }
        if(node.left is not None)
        {
            lifo.push(node.left);
        }
    }
}

For postorder traversal you simply swap the order you push the left and right subtrees onto the stack.