# Building Expression tree from Prefix Expression

Given a character array a[] representing a prefix expression. The task is to build an Expression Tree for the expression and then print the infix and postfix expression of the built tree.

Examples:

Input: a[] = “*+ab-cd”
Output: The Infix expression is:
a + b * c – d
The Postfix expression is:
a b + c d – *

Input: a[] = “+ab”
Output: The Infix expression is:
a + b
The Postfix expression is:
a b +

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: If the character is an operand i.e. X then it’ll be the leaf node of the required tree as all the operands are at the leaf in an expression tree. Else if the character is an operator and of the form OP X Y then it’ll be an internal node with left child as the expressionTree(X) and right child as the expressionTree(Y) which can be solved using a recursive function.

Below is the implementation of the above approach:

 `// C program to construct an expression tree  ` `// from prefix expression ` `#include ` `#include ` ` `  `// Represents a node of the required tree ` `typedef` `struct` `node { ` `    ``char` `data; ` `    ``struct` `node *left, *right; ` `} node; ` ` `  `// Function to recursively build the expression tree ` `char``* add(node** p, ``char``* a) ` `{ ` ` `  `    ``// If its the end of the expression ` `    ``if` `(*a == ``'\0'``) ` `        ``return` `'\0'``; ` ` `  `    ``while` `(1) { ` `        ``char``* q = ``"null"``; ` `        ``if` `(*p == NULL) { ` ` `  `            ``// Create a node with *a as the data and ` `            ``// both the children set to null ` `            ``node* nn = (node*)``malloc``(``sizeof``(node)); ` `            ``nn->data = *a; ` `            ``nn->left = NULL; ` `            ``nn->right = NULL; ` `            ``*p = nn; ` `        ``} ` `        ``else` `{ ` ` `  `            ``// If the character is an operand ` `            ``if` `(*a >= ``'a'` `&& *a <= ``'z'``) { ` `                ``return` `a; ` `            ``} ` ` `  `            ``// Build the left sub-tree ` `            ``q = add(&(*p)->left, a + 1); ` ` `  `            ``// Build the right sub-tree ` `            ``q = add(&(*p)->right, q + 1); ` ` `  `            ``return` `q; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to print the infix expression for the tree ` `void` `inr(node* p) ``// recursion ` `{ ` `    ``if` `(p == NULL) { ` `        ``return``; ` `    ``} ` `    ``else` `{ ` `        ``inr(p->left); ` `        ``printf``(``"%c "``, p->data); ` `        ``inr(p->right); ` `    ``} ` `} ` ` `  `// Function to print the postfix expression for the tree ` `void` `postr(node* p) ` `{ ` `    ``if` `(p == NULL) { ` `        ``return``; ` `    ``} ` `    ``else` `{ ` `        ``postr(p->left); ` `        ``postr(p->right); ` `        ``printf``(``"%c "``, p->data); ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``node* s = NULL; ` `    ``char` `a[] = ``"*+ab-cd"``; ` `    ``add(&s, a); ` `    ``printf``(``"The Infix expression is:\n "``); ` `    ``inr(s); ` `    ``printf``(``"\n"``); ` `    ``printf``(``"The Postfix expression is:\n "``); ` `    ``postr(s); ` `    ``return` `0; ` `} `

Output:

```The Infix expression is:
a + b * c - d
The Postfix expression is:
a b + c d - *
```

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