The use of a tree structure involves traversing or stepping through the elements or nodes of the tree. This page shows how to traverse a binary tree, which we can extend to trees having more than two descendants at each node. Many problems can be modeled by a tree. For example, in chess, the first move can be represented by the root or starting node of a tree; the next move by an opponent player, by the descendent nodes of the root. This decomposition can continue for many levels. Thus, a level in the tree hierarchy represents the possible moves available to one player; and the next level, the possible moves of the opponent player. Each level represents the choices available to a given player. Traversing the tree involves: from a given start node a player looks-ahead at its descendent nodes (the possible moves), from each of these descendant nodes the player looks-ahead at their descendants (possible responding moves of the opponent player), and so on, continuing to look ahead (planning) to cover as many levels as feasible. Based on the look-ahead information (which gets better the further the look-ahead goes), the player chooses a descendant from the given start node.

If the node of a Binary tree is defined as below:

struct Node{ Node * lptr; // Pointer to Left subtree int data; Node *rptr; // Pointer to Right subtree };

write functions which will traverse the tree in in-Order, Pre-Order and Post-Order.

In earlier questions related to tree, “Counting number of leaf nodes” and “Has-Path sum problem“, I have written that tree algorithms are better written using recursive algorithms. These traversal algorithms also are easy if we write them recursively.

For demo purpose consider the below tree.

The three traversals (pre, in and post order) are named so because of the positioning of root node w.r.t the left and Right sub tree. In pre-order traversal, you visit the root node, then you visit the left subtree in pre-order traversal and then the right subtree in pre-order. In In-Order, we traverse the left subtree in In-Order then we visit the root and finally the Right sub-tree in In-Order, Similarly in Post-Order traversal.

Lets start our traversals:

**Pre-order traversal**

Visit the root node first (pre)

**Algorithm:**

1. Visit the root (we will print it when we visit to show the order of visiting) 2. Traverse the left subtree in pre-order 3. Traverse the right subtree in pre-order

**Code:**

/* Print Pre-order traversal of the tree */ void preOrder(node* r){ if(r){ cout << r->data << " "; preOrder(r->left); preOrder(r->right); } }

**Output:**

10 5 4 1 8 30 40

The root node comes before the left and right subtree (for every node of the tree)

**In-order traversal**

Visit the root node in between the left and right node (in)

**Algorithm:**

1. Traverse the left subtree in in-order 2. Visit the root (we will print it when we visit to show the order of visiting) 3. Traverse the right subtree in in-order

**Code:**

/* Print In-order traversal of the tree */ void inOrder(node* r){ if(r){ inOrder(r->left); cout << r->data << " "; inOrder(r->right); } }

**Output:**

1 4 5 8 10 30 40

You may want to observe in the output that root lies in between the left and right traversals.

**Post-order traversal**

Visit the root node after (post) visiting the left and right subtree.

**Algorithm:**

1. Traverse the left subtree in in-order 2. Traverse the right subtree in in-order 3. Visit the root (we will print it when we visit to show the order of visiting)

**Code:**

/* Print Post-order traversal of the tree */ void postOrder(node* r){ if(r){ postOrder(r->left); postOrder(r->right); cout << r->data << " "; } }

**Output:**

1 4 8 5 40 30 10

Source: http://www.ritambhara.in/basic-tree-traversals-preorder-inorder-postorder/

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Last modified: Tuesday, October 15, 2019, 12:08 PM