Tree Traversals (Inorder, Preorder and Postorder)

Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. Following are the generally used ways for traversing trees.

Example Tree

Depth First Traversals:
(a) Inorder (Left, Root, Right) : 4 2 5 1 3
(b) Preorder (Root, Left, Right) : 1 2 4 5 3
(c) Postorder (Left, Right, Root) : 4 5 2 3 1
 

Breadth First or Level Order Traversal : 1 2 3 4 5
Please see this post for Breadth First Traversal.

Inorder Traversal (Practice):

Algorithm Inorder(tree)
   1. Traverse the left subtree, i.e., call Inorder(left-subtree)
   2. Visit the root.
   3. Traverse the right subtree, i.e., call Inorder(right-subtree)
 

Uses of Inorder
In case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder traversal s reversed can be used.
Example: Inorder traversal for the above-given figure is 4 2 5 1 3.

Preorder Traversal (Practice):

Algorithm Preorder(tree)
   1. Visit the root.
   2. Traverse the left subtree, i.e., call Preorder(left-subtree)
   3. Traverse the right subtree, i.e., call Preorder(right-subtree) 

Uses of Preorder
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expression on of an expression tree. Please see http://en.wikipedia.org/wiki/Polish_notation to know why prefix expressions are useful.
Example: Preorder traversal for the above given figure is 1 2 4 5 3.

Postorder Traversal (Practice):

Algorithm Postorder(tree)
   1. Traverse the left subtree, i.e., call Postorder(left-subtree)
   2. Traverse the right subtree, i.e., call Postorder(right-subtree)
   3. Visit the root.

Uses of Postorder
Postorder traversal is used to delete the tree. Please see the question for deletion of tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree. Please see http://en.wikipedia.org/wiki/Reverse_Polish_notation to for the usage of postfix expression.

Example: Postorder traversal for the above given figure is 4 5 2 3 1.

# Python program to for tree traversals 

# A class that represents an individual node in a 
# Binary Tree 
class Node: 
	def __init__(self,key): 
		self.left = None
		self.right = None
		self.val = key 


# A function to do inorder tree traversal 
def printInorder(root): 

	if root: 

		# First recur on left child 
		printInorder(root.left) 

		# then print the data of node 
		print(root.val), 

		# now recur on right child 
		printInorder(root.right) 



# A function to do postorder tree traversal 
def printPostorder(root): 

	if root: 

		# First recur on left child 
		printPostorder(root.left) 

		# the recur on right child 
		printPostorder(root.right) 

		# now print the data of node 
		print(root.val), 


# A function to do preorder tree traversal 
def printPreorder(root): 

	if root: 

		# First print the data of node 
		print(root.val), 

		# Then recur on left child 
		printPreorder(root.left) 

		# Finally recur on right child 
		printPreorder(root.right) 


# Driver code 
root = Node(1) 
root.left	 = Node(2) 
root.right	 = Node(3) 
root.left.left = Node(4) 
root.left.right = Node(5) 
print "Preorder traversal of binary tree is"
printPreorder(root) 

print "\nInorder traversal of binary tree is"
printInorder(root) 

print "\nPostorder traversal of binary tree is"
printPostorder(root) 

Output:
Preorder traversal of binary tree is
1 2 4 5 3 
Inorder traversal of binary tree is
4 2 5 1 3 
Postorder traversal of binary tree is
4 5 2 3 1

One more example:

Time Complexity: O(n)
Let us see different corner cases.
Complexity function T(n) — for all problem where tree traversal is involved — can be defined as:

T(n) = T(k) + T(n – k – 1) + c

Where k is the number of nodes on one side of root and n-k-1 on the other side.

Let’s do an analysis of boundary conditions

Case 1: Skewed tree (One of the subtrees is empty and other subtree is non-empty )

k is 0 in this case.
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c

…………………………………………
………………………………………….
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0) + (n)c

Value of T(0) will be some constant say d. (traversing a empty tree will take some constants time)

T(n) = n(c+d)
T(n) = Θ(n) (Theta of n)

Case 2: Both left and right subtrees have equal number of nodes.

T(n) = 2T(|n/2|) + c

This recursive function is in the standard form (T(n) = aT(n/b) + (-)(n) ) for master method http://en.wikipedia.org/wiki/Master_theorem. If we solve it by master method we get (-)(n)

Auxiliary Space : If we don’t consider size of stack for function calls then O(1) otherwise O(n).

Reference:

https://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/