Tree traversal

Template:Graph search algorithm In computer science, tree-traversal refers to the process of visiting (examining and/or updating) each node in a tree data structure, exactly once, in a systematic way. Such traversals are classified by the order in which the nodes are visited. The following algorithms are described for a binary tree, but they may be generalized to other trees as well.

Traversal

Compared to linear data structures like linked lists and one dimensional arrays, which have only one logical means of traversal, tree structures can be traversed in many different ways. Starting at the root of a binary tree, there are three main steps that can be performed and the order in which they are performed defines the traversal type. These steps (in no particular order) are: performing an action on the current node (referred to as "visiting" the node), traversing to the left child node, and traversing to the right child node. Thus the process is most easily described through recursion. The names given for particular style of traversal came from the position of root element with regard to the left and right nodes. Imagine that the left and right nodes are constant in space, then the root node could be placed to the left of the left node (pre-order), between the left and right node (in-order), or to the right of the right node (post-order).

Depth-first Traversal

To traverse a non-empty binary tree in preorder, perform the following operations recursively at each node, starting with the root node:

Visit the root.
Traverse the left subtree.
Traverse the right subtree.

To traverse a non-empty binary tree in inorder (symmetric), perform the following operations recursively at each node:

Traverse the left subtree.
Visit the root.
Traverse the right subtree.

To traverse a non-empty binary tree in postorder, perform the following operations recursively at each node:

Traverse the left subtree.
Traverse the right subtree.
Visit the root.

For more information, see Depth-first traversal.

Breadth-first Traversal

Finally, trees can also be traversed in level-order, where we visit every node on a level before going to a lower level. This is also called Breadth-first traversal.

Example

In this binary search tree

Preorder traversal sequence: F, B, A, D, C, E, G, I, H (root, left, right)
Inorder traversal sequence: A, B, C, D, E, F, G, H, I (left, root, right); note how this produces a sorted sequence
Postorder traversal sequence: A, C, E, D, B, H, I, G, F (left, right, root)
Level-order traversal sequence: F, B, G, A, D, I, C, E, H

pre-order

in-order

post-order

level-order

push F
pop F
push G B
pop B
push D A
pop A
pop D
push E C
pop C
pop E
pop G
push I
pop I
push H
pop H

push F B A
pop A
pop B
push D C
pop C
pop D
push E
pop E
push F
pop F
push G
pop G
push I H
pop H
pop I

push F B A
pop A
push D C
pop C
push E
pop E
pop D
pop B
push G I H
pop H
pop I
pop G
pop F

push F B G
pop F
push A D
pop B
push I
pop G
pop A
push C E
pop D
push H
pop I
pop C
pop E
pop H

Sample implementations

preorder(node)
  print node.value
  if node.left ≠ null then preorder(node.left) 
  if node.right ≠ null then preorder(node.right)

inorder(node)
  if node.left  ≠ null then inorder(node.left)
  print node.value
  if node.right ≠ null then inorder(node.right)

postorder(node)
  if node.left  ≠ null then postorder(node.left)
  if node.right ≠ null then postorder(node.right)
  print node.value

All sample implementations will require call stack space proportional to the height of the tree. In a poorly balanced tree, this can be quite considerable.

We can remove the stack requirement by maintaining parent pointers in each node, or by threading the tree. In the case of using threads, this will allow for greatly improved inorder traversal, although retrieving the parent node required for preorder and postorder traversal will be slower than a simple stack based algorithm.

To traverse a threaded tree inorder, we could do something like this:

inorder(node)
  while hasleftchild(node) do
    node = node.left
  do
    visit(node)
    if (hasrightchild(node)) then
      node = node.right
      while hasleftchild(node) do
        node = node.left
    else
      while node.parent ≠ null and node = node.parent.right
        node = node.parent
      node = node.parent
  while node ≠ null

Note that a threaded binary tree will provide a means of determining whether a pointer is a child, or a thread. See threaded binary trees for more information.

Queue-based level order traversal

Also, listed below is pseudocode for a simple queue based level order traversal, and will require space proportional to the maximum number of nodes at a given depth. This can be as much as the total number of nodes / 2. A more space-efficient approach for this type of traversal can be implemented using an iterative deepening depth-first search.

levelorder(root) 
  q = empty queue
  q.enqueue(root)
  while not q.empty do
    node := q.dequeue()
    visit(node)
    if node.left ≠ null
      q.enqueue(node.left)
    if node.right ≠ null
      q.enqueue(node.right)

Uses

Inorder traversal

It is particularly common to use an inorder traversal on a binary search tree because this will return values from the underlying set in order, according to the comparator that set up the binary search tree (hence the name).

To see why this is the case, note that if n is a node in a binary search tree, then everything in n 's left subtree is less than n, and everything in n 's right subtree is greater than or equal to n. Thus, if we visit the left subtree in order, using a recursive call, and then visit n, and then visit the right subtree in order, we have visited the entire subtree rooted at n in order. We can assume the recursive calls correctly visit the subtrees in order using the mathematical principle of structural induction. Traversing in reverse inorder similarly gives the values in decreasing order.

Preorder traversal

Traversing a tree in preorder while inserting the values into a new tree is common way of making a complete copy of a binary search tree.

One can also use preorder traversals to get a prefix expression (Polish notation) from expression trees: traverse the expression tree preorderly. To calculate the value of such an expression: scan from right to left, placing the elements in a stack. Each time we find an operator, we replace the two top symbols of the stack with the result of applying the operator to those elements. For instance, the expression ∗ + 2 3 4, which in infix notation is (2 + 3) ∗ 4, would be evaluated like this:

**Using prefix traversal to evaluate an expression tree**
Expression (remaining)	Stack
∗ + 2 3 4	<empty>
∗ + 2 3	4
∗ + 2	3 4
∗ +	2 3 4
∗	5 4
Answer	20

Functional traversal

We could perform the same traversals in a functional language like Haskell using code similar to this:

 data Tree a = Nil | Node (Tree a) a (Tree a)

 preorder Nil = []
 preorder (Node left x right) = [x] ++ (preorder left) ++ (preorder right)

 postorder Nil = []
 postorder (Node left x right) = (postorder left) ++ (postorder right) ++ [x]

 inorder Nil = []
 inorder (Node left x right) = (inorder left) ++ [x] ++ (inorder right)

 levelorder t = step [t]
     where
           step [] = []
           step ts = concatMap elements ts ++ step (concatMap subtrees ts)
           elements Nil = []
           elements (Node left x right) = [x]
           subtrees Nil = []
           subtrees (Node left x right) = [left, right]

Iterative Traversal

All the above recursive algorithms require stack space proportional to the depth of the tree. Recursive traversal may be converted into an iterative one using various well-known methods.

A sample is shown here for postorder traversal using a visited flag:

iterativePostorder(rootNode)
  nodeStack.push(rootNode)
  while (! nodeStack.empty())
    currNode = nodeStack.peek()
    if ((currNode.left != null) and (currNode.left.visited == false))
      nodeStack.push(currNode.left)
    else 
      if ((currNode.right != null) and (currNode.right.visited == false))
        nodeStack.push(currNode.right)
      else
        print currNode.value
        currNode.visited := true
        nodeStack.pop()

In this case each node is required to keep an additional "visited" flag, other than usual information (value, left-child-reference, right-child-reference).

Another example is preorder traversal without using a visited flag (Java):

public void iterativePreorder(Node root) {
   Stack nodes = new Stack();
   nodes.push(root);
   Node currentNode;
   while (! nodes.isEmpty()){
      currentNode = nodes.pop();
      Node right = currentNode.right();
      if (right != null)
         nodes.push(right);
      Node left = currentNode.left();
      if (left != null)
         nodes.push(left);      
      System.out.println("Node data: "+currentNode.data);
   }
}

If each node holds reference to the parent node, then iterative traversal is possible without a stack or "visited" flag.

References

Dale, Nell. Lilly, Susan D. "Pascal Plus Data Structures". D. C. Heath and Company. Lexington, MA. 1995. Fourth Edition.
Drozdek, Adam. "Data Structures and Algorithms in C++". Brook/Cole. Pacific Grove, CA. 2001. Second edition.
http://www.math.northwestern.edu/~mlerma/courses/cs310-05s/notes/dm-treetran

External links