# Tree Traversals (Inorder, Preorder and Postorder)

• Difficulty Level : Easy
• Last Updated : 16 Jul, 2022

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.

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

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 the 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: In order 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 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 the deletion of a 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 for the usage of postfix expression.

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

## C++

 `// C++ program for different tree traversals``#include ``using` `namespace` `std;`` ` `/* A binary tree node has data, pointer to left child``and a pointer to right child */``struct` `Node {``    ``int` `data;``    ``struct` `Node *left, *right;``};`` ` `//Utility function to create a new tree node``Node* newNode(``int` `data)``{``    ``Node* temp = ``new` `Node;``    ``temp->data = data;``    ``temp->left = temp->right = NULL;``    ``return` `temp;``}`` ` `/* Given a binary tree, print its nodes according to the``"bottom-up" postorder traversal. */``void` `printPostorder(``struct` `Node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``// first recur on left subtree``    ``printPostorder(node->left);`` ` `    ``// then recur on right subtree``    ``printPostorder(node->right);`` ` `    ``// now deal with the node``    ``cout << node->data << ``" "``;``}`` ` `/* Given a binary tree, print its nodes in inorder*/``void` `printInorder(``struct` `Node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``/* first recur on left child */``    ``printInorder(node->left);`` ` `    ``/* then print the data of node */``    ``cout << node->data << ``" "``;`` ` `    ``/* now recur on right child */``    ``printInorder(node->right);``}`` ` `/* Given a binary tree, print its nodes in preorder*/``void` `printPreorder(``struct` `Node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``/* first print data of node */``    ``cout << node->data << ``" "``;`` ` `    ``/* then recur on left subtree */``    ``printPreorder(node->left);`` ` `    ``/* now recur on right subtree */``    ``printPreorder(node->right);``}`` ` `/* Driver program to test above functions*/``int` `main()``{``    ``struct` `Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);`` ` `    ``cout << ``"\nPreorder traversal of binary tree is \n"``;``    ``printPreorder(root);`` ` `    ``cout << ``"\nInorder traversal of binary tree is \n"``;``    ``printInorder(root);`` ` `    ``cout << ``"\nPostorder traversal of binary tree is \n"``;``    ``printPostorder(root);`` ` `    ``return` `0;``}`

## C

 `// C program for different tree traversals``#include ``#include `` ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``struct` `node {``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};`` ` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node``        ``= (``struct` `node*)``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;`` ` `    ``return` `(node);``}`` ` `/* Given a binary tree, print its nodes according to the``  ``"bottom-up" postorder traversal. */``void` `printPostorder(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``// first recur on left subtree``    ``printPostorder(node->left);`` ` `    ``// then recur on right subtree``    ``printPostorder(node->right);`` ` `    ``// now deal with the node``    ``printf``(``"%d "``, node->data);``}`` ` `/* Given a binary tree, print its nodes in inorder*/``void` `printInorder(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``/* first recur on left child */``    ``printInorder(node->left);`` ` `    ``/* then print the data of node */``    ``printf``(``"%d "``, node->data);`` ` `    ``/* now recur on right child */``    ``printInorder(node->right);``}`` ` `/* Given a binary tree, print its nodes in preorder*/``void` `printPreorder(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return``;`` ` `    ``/* first print data of node */``    ``printf``(``"%d "``, node->data);`` ` `    ``/* then recur on left subtree */``    ``printPreorder(node->left);`` ` `    ``/* now recur on right subtree */``    ``printPreorder(node->right);``}`` ` `/* Driver program to test above functions*/``int` `main()``{``    ``struct` `node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);`` ` `    ``printf``(``"\nPreorder traversal of binary tree is \n"``);``    ``printPreorder(root);`` ` `    ``printf``(``"\nInorder traversal of binary tree is \n"``);``    ``printInorder(root);`` ` `    ``printf``(``"\nPostorder traversal of binary tree is \n"``);``    ``printPostorder(root);`` ` `    ``getchar``();``    ``return` `0;``}`

## Python

 `# 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)`

## Java

 `// Java program for different tree traversals`` ` `/* Class containing left and right child of current``   ``node and key value*/``class` `Node {``    ``int` `key;``    ``Node left, right;`` ` `    ``public` `Node(``int` `item)``    ``{``        ``key = item;``        ``left = right = ``null``;``    ``}``}`` ` `class` `BinaryTree {``    ``// Root of Binary Tree``    ``Node root;`` ` `    ``BinaryTree() { root = ``null``; }`` ` `    ``/* Given a binary tree, print its nodes according to the``      ``"bottom-up" postorder traversal. */``    ``void` `printPostorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``// first recur on left subtree``        ``printPostorder(node.left);`` ` `        ``// then recur on right subtree``        ``printPostorder(node.right);`` ` `        ``// now deal with the node``        ``System.out.print(node.key + ``" "``);``    ``}`` ` `    ``/* Given a binary tree, print its nodes in inorder*/``    ``void` `printInorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``/* first recur on left child */``        ``printInorder(node.left);`` ` `        ``/* then print the data of node */``        ``System.out.print(node.key + ``" "``);`` ` `        ``/* now recur on right child */``        ``printInorder(node.right);``    ``}`` ` `    ``/* Given a binary tree, print its nodes in preorder*/``    ``void` `printPreorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``/* first print data of node */``        ``System.out.print(node.key + ``" "``);`` ` `        ``/* then recur on left subtree */``        ``printPreorder(node.left);`` ` `        ``/* now recur on right subtree */``        ``printPreorder(node.right);``    ``}`` ` `    ``// Wrappers over above recursive functions``    ``void` `printPostorder() { printPostorder(root); }``    ``void` `printInorder() { printInorder(root); }``    ``void` `printPreorder() { printPreorder(root); }`` ` `    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(``1``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``3``);``        ``tree.root.left.left = ``new` `Node(``4``);``        ``tree.root.left.right = ``new` `Node(``5``);`` ` `        ``System.out.println(``            ``"Preorder traversal of binary tree is "``);``        ``tree.printPreorder();`` ` `        ``System.out.println(``            ``"\nInorder traversal of binary tree is "``);``        ``tree.printInorder();`` ` `        ``System.out.println(``            ``"\nPostorder traversal of binary tree is "``);``        ``tree.printPostorder();``    ``}``}`

## C#

 `// C# program for different``// tree traversals``using` `System;`` ` `/* Class containing left and``right child of current``node and key value*/``class` `Node {``    ``public` `int` `key;``    ``public` `Node left, right;`` ` `    ``public` `Node(``int` `item)``    ``{``        ``key = item;``        ``left = right = ``null``;``    ``}``}`` ` `class` `BinaryTree {``    ``// Root of Binary Tree``    ``Node root;`` ` `    ``BinaryTree() { root = ``null``; }`` ` `    ``/* Given a binary tree, print``       ``its nodes according to the``       ``"bottom-up" postorder traversal. */``    ``void` `printPostorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``// first recur on left subtree``        ``printPostorder(node.left);`` ` `        ``// then recur on right subtree``        ``printPostorder(node.right);`` ` `        ``// now deal with the node``        ``Console.Write(node.key + ``" "``);``    ``}`` ` `    ``/* Given a binary tree, print``       ``its nodes in inorder*/``    ``void` `printInorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``/* first recur on left child */``        ``printInorder(node.left);`` ` `        ``/* then print the data of node */``        ``Console.Write(node.key + ``" "``);`` ` `        ``/* now recur on right child */``        ``printInorder(node.right);``    ``}`` ` `    ``/* Given a binary tree, print``       ``its nodes in preorder*/``    ``void` `printPreorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``/* first print data of node */``        ``Console.Write(node.key + ``" "``);`` ` `        ``/* then recur on left subtree */``        ``printPreorder(node.left);`` ` `        ``/* now recur on right subtree */``        ``printPreorder(node.right);``    ``}`` ` `    ``// Wrappers over above recursive functions``    ``void` `printPostorder() { printPostorder(root); }``    ``void` `printInorder() { printInorder(root); }``    ``void` `printPreorder() { printPreorder(root); }`` ` `    ``// Driver Code``    ``static` `public` `void` `Main(String[] args)``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(1);``        ``tree.root.left = ``new` `Node(2);``        ``tree.root.right = ``new` `Node(3);``        ``tree.root.left.left = ``new` `Node(4);``        ``tree.root.left.right = ``new` `Node(5);`` ` `        ``Console.WriteLine(``"Preorder traversal "``                          ``+ ``"of binary tree is "``);``        ``tree.printPreorder();`` ` `        ``Console.WriteLine(``"\nInorder traversal "``                          ``+ ``"of binary tree is "``);``        ``tree.printInorder();`` ` `        ``Console.WriteLine(``"\nPostorder traversal "``                          ``+ ``"of binary tree is "``);``        ``tree.printPostorder();``    ``}``}`` ` `// This code is contributed by Arnab Kundu`

## Javascript

 ``

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 problems 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 the 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 another 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 an 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 an 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 the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree.

The height of the skewed tree is n (no. of elements) so the worst space complexity is O(n) and height is (Log n) for the balanced tree so the best space complexity is O(Log n).

My Personal Notes arrow_drop_up