# Program to convert Infix notation to Expression Tree

Given a string representing infix notation. The task is to convert it to an expression tree.
Expression Tree is a binary tree where the operands are represented by leaf nodes and operators are represented by intermediate nodes. No node can have a single child.

Construction of Expression tree

The algorithm follows a combination of shunting yard along with postfix-to-expression tree conversion.

Consider the below line:

```((s[i]!='^' && p[stC.top()]>=p[s[i]]) ||
(s[i]=='^' && p[stC.top()]>p[s[i]])))```

You might remember that unlike ‘+’, ‘-‘, ‘*’ and ‘/’; ‘^’ is right associative.
In simpler terms, a^b^c is a^(b^c) not (a^b)^c. So it must be evaluated from the right.

Now lets take a look at how does the algorithm work,(Take a quick glance at the code to get a better idea of the variables used)

```Let us have an expression s = ((a+b)*c-e*f)
currently both the stacks are empty-
(we'll use C to denote the char stack and N for node stack)
s[0] = '('            ((a+b)*c-e*f)
^
C|(|, N| |

s[1] = '('           ((a+b)*c-e*f)
^
|(|
C|(|, N| |

s[2] = 'a'            ((a+b)*c-e*f)
^
|(|
C|(|, N|a|

s[3] = '+'            ((a+b)*c-e*f)
^
|+|
|(|
C|(|, N|a|

s[4] = 'b'             ((a+b)*c-e*f)
^
|+|
|(|   |b|
C|(|, N|a|

s[5] = ')'             ((a+b)*c-e*f)
^
|+|            t = '+'         +
|(|   |b|  ->  t1= 'b'        / \    ->
C|(|, N|a|      t2= 'a'       a   b      C|(|, N|+|

s[6] = '*'              ((a+b)*c-e*f)
^
|*|
C|(|, N|+|

s[7] = 'c'              ((a+b)*c-e*f)
^
|*|   |c|
C|(|, N|+|

s[8] = '-'   ((a+b)*c-e*f)              now (C.top(*)>s[8](-))
^    t = '*'        *
|*|   |c|            t1 = c        / \  ->    |-|
C|(|, N|+|            t2 = +       +   c      C|(|, N|*|
/ \
a   b
s[9] = 'e'            ((a+b)*c-e*f)
^
|-|   |e|
C|(|, N|*|

s[10] = '*'            ((a+b)*c-e*f)      now (C.top(-)>s[10](*))
^
|*|
|-|   |e|
C|(|, N|*|

s[11] = 'f'             ((a+b)*c-e*f)
^
|*|   |f|
|-|   |e|
C|(|, N|*|

s[12] = ')'             ((a+b)*c-e*f)
1>                               ^
|*|   |f|         t = '*'          *
|-|   |e|  ->     t1= 'f'  ->     / \  ->   |-|   |*|
C|(|, N|*|         t2= 'e'        e   f     C|(|, N|*|

2>
t = '-'           -
|-|   |*|  ->     t1= '*'  ->     /   \  ->
C|(|, N|*|         t2= '*'        *     *     C| |, N|-|
/ \   / \
+   c  e  f
/ \
a   b
now make (-) the root of the tree```

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Tree Structure` `typedef` `struct` `node ` `{` `    ``char` `data;` `    ``struct` `node *left, *right;` `} * nptr;`   `// Function to create new node` `nptr newNode(``char` `c)` `{` `    ``nptr n = ``new` `node;` `    ``n->data = c;` `    ``n->left = n->right = nullptr;` `    ``return` `n;` `}`   `// Function to build Expression Tree` `nptr build(string& s)` `{`   `    ``// Stack to hold nodes` `    ``stack stN;`   `    ``// Stack to hold chars` `    ``stack<``char``> stC;` `    ``nptr t, t1, t2;`   `    ``// Prioritising the operators` `    ``int` `p[123] = { 0 };` `    ``p[``'+'``] = p[``'-'``] = 1, p[``'/'``] = p[``'*'``] = 2, p[``'^'``] = 3,` `    ``p[``')'``] = 0;`   `    ``for` `(``int` `i = 0; i < s.length(); i++) ` `    ``{` `        ``if` `(s[i] == ``'('``) {`   `            ``// Push '(' in char stack` `            ``stC.push(s[i]);` `        ``}`   `        ``// Push the operands in node stack` `        ``else` `if` `(``isalpha``(s[i])) ` `        ``{` `            ``t = newNode(s[i]);` `            ``stN.push(t);` `        ``}` `        ``else` `if` `(p[s[i]] > 0) ` `        ``{` `            ``// If an operator with lower or` `            ``// same associativity appears` `            ``while` `(` `                ``!stC.empty() && stC.top() != ``'('` `                ``&& ((s[i] != ``'^'` `&& p[stC.top()] >= p[s[i]])` `                    ``|| (s[i] == ``'^'` `                        ``&& p[stC.top()] > p[s[i]]))) ` `            ``{`   `                ``// Get and remove the top element` `                ``// from the character stack` `                ``t = newNode(stC.top());` `                ``stC.pop();`   `                ``// Get and remove the top element` `                ``// from the node stack` `                ``t1 = stN.top();` `                ``stN.pop();`   `                ``// Get and remove the currently top` `                ``// element from the node stack` `                ``t2 = stN.top();` `                ``stN.pop();`   `                ``// Update the tree` `                ``t->left = t2;` `                ``t->right = t1;`   `                ``// Push the node to the node stack` `                ``stN.push(t);` `            ``}`   `            ``// Push s[i] to char stack` `            ``stC.push(s[i]);` `        ``}` `        ``else` `if` `(s[i] == ``')'``) {` `            ``while` `(!stC.empty() && stC.top() != ``'('``) ` `            ``{` `                ``t = newNode(stC.top());` `                ``stC.pop();` `                ``t1 = stN.top();` `                ``stN.pop();` `                ``t2 = stN.top();` `                ``stN.pop();` `                ``t->left = t2;` `                ``t->right = t1;` `                ``stN.push(t);` `            ``}` `            ``stC.pop();` `        ``}` `    ``}` `    ``t = stN.top();` `    ``return` `t;` `}`   `// Function to print the post order` `// traversal of the tree` `void` `postorder(nptr root)` `{` `    ``if` `(root) ` `    ``{` `        ``postorder(root->left);` `        ``postorder(root->right);` `        ``cout << root->data;` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"(a^b^(c/d/e-f)^(x*y-m*n))"``;` `    ``s = ``"("` `+ s;` `    ``s += ``")"``;` `    ``nptr root = build(s);` `  `  `    ``// Function call` `    ``postorder(root);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG` `{`   `// Tree Structure` `static` `class` `nptr ` `{` `    ``char` `data;` `    ``nptr left, right;` `} ;`   `// Function to create new node` `static` `nptr newNode(``char` `c)` `{` `    ``nptr n = ``new` `nptr();` `    ``n.data = c;` `    ``n.left = n.right = ``null``;` `    ``return` `n;` `}`   `// Function to build Expression Tree` `static` `nptr build(String s)` `{`   `    ``// Stack to hold nodes` `    ``Stack stN = ``new` `Stack<>();`   `    ``// Stack to hold chars` `    ``Stack stC = ``new` `Stack<>();` `    ``nptr t, t1, t2;`   `    ``// Prioritising the operators` `    ``int` `[]p = ``new` `int``[``123``];` `    ``p[``'+'``] = p[``'-'``] = ``1``;` `    ``p[``'/'``] = p[``'*'``] = ``2``;` `    ``p[``'^'``] = ``3``;` `    ``p[``')'``] = ``0``;`   `    ``for` `(``int` `i = ``0``; i < s.length(); i++) ` `    ``{` `        ``if` `(s.charAt(i) == ``'('``) {`   `            ``// Push '(' in char stack` `            ``stC.add(s.charAt(i));` `        ``}`   `        ``// Push the operands in node stack` `        ``else` `if` `(Character.isAlphabetic(s.charAt(i))) ` `        ``{` `            ``t = newNode(s.charAt(i));` `            ``stN.add(t);` `        ``}` `        ``else` `if` `(p[s.charAt(i)] > ``0``) ` `        ``{` `          `  `            ``// If an operator with lower or` `            ``// same associativity appears` `            ``while` `(` `                ``!stC.isEmpty() && stC.peek() != ``'('` `                ``&& ((s.charAt(i) != ``'^'` `&& p[stC.peek()] >= p[s.charAt(i)])` `                    ``|| (s.charAt(i) == ``'^'` `                        ``&& p[stC.peek()] > p[s.charAt(i)]))) ` `            ``{`   `                ``// Get and remove the top element` `                ``// from the character stack` `                ``t = newNode(stC.peek());` `                ``stC.pop();`   `                ``// Get and remove the top element` `                ``// from the node stack` `                ``t1 = stN.peek();` `                ``stN.pop();`   `                ``// Get and remove the currently top` `                ``// element from the node stack` `                ``t2 = stN.peek();` `                ``stN.pop();`   `                ``// Update the tree` `                ``t.left = t2;` `                ``t.right = t1;`   `                ``// Push the node to the node stack` `                ``stN.add(t);` `            ``}`   `            ``// Push s[i] to char stack` `            ``stC.push(s.charAt(i));` `        ``}` `        ``else` `if` `(s.charAt(i) == ``')'``) {` `            ``while` `(!stC.isEmpty() && stC.peek() != ``'('``) ` `            ``{` `                ``t = newNode(stC.peek());` `                ``stC.pop();` `                ``t1 = stN.peek();` `                ``stN.pop();` `                ``t2 = stN.peek();` `                ``stN.pop();` `                ``t.left = t2;` `                ``t.right = t1;` `                ``stN.add(t);` `            ``}` `            ``stC.pop();` `        ``}` `    ``}` `    ``t = stN.peek();` `    ``return` `t;` `}`   `// Function to print the post order` `// traversal of the tree` `static` `void` `postorder(nptr root)` `{` `    ``if` `(root != ``null``) ` `    ``{` `        ``postorder(root.left);` `        ``postorder(root.right);` `        ``System.out.print(root.data);` `    ``}` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String s = ``"(a^b^(c/d/e-f)^(x*y-m*n))"``;` `    ``s = ``"("` `+ s;` `    ``s += ``")"``;` `    ``nptr root = build(s);` `  `  `    ``// Function call` `    ``postorder(root);` `}` `}`   `// This code is contributed by aashish1995 `

## Python3

 `# Python3 implementation of the approach`   `# Tree Structure` `class` `nptr:` `  `  `    ``# Constructor to set the data of` `    ``# the newly created tree node` `    ``def` `__init__(``self``, c):` `        ``self``.data ``=` `c` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to create new node` `def` `newNode(c):` `    ``n ``=` `nptr(c)` `    ``return` `n`   `# Function to build Expression Tree` `def` `build(s):` `  `  `    ``# Stack to hold nodes` `    ``stN ``=` `[]`   `    ``# Stack to hold chars` `    ``stC ``=` `[]`   `    ``# Prioritising the operators` `    ``p ``=` `[``0``]``*``(``123``)` `    ``p[``ord``(``'+'``)] ``=` `p[``ord``(``'-'``)] ``=` `1` `    ``p[``ord``(``'/'``)] ``=` `p[``ord``(``'*'``)] ``=` `2` `    ``p[``ord``(``'^'``)] ``=` `3` `    ``p[``ord``(``')'``)] ``=` `0`   `    ``for` `i ``in` `range``(``len``(s)):` `        ``if` `(s[i] ``=``=` `'('``):` `            ``# Push '(' in char stack` `            ``stC.append(s[i])`   `        ``# Push the operands in node stack` `        ``elif` `(s[i].isalpha()):` `            ``t ``=` `newNode(s[i])` `            ``stN.append(t)` `        ``elif` `(p[``ord``(s[i])] > ``0``):` `          `  `            ``# If an operator with lower or` `            ``# same associativity appears` `            ``while` `(``len``(stC) !``=` `0` `and` `stC[``-``1``] !``=` `'('` `and` `((s[i] !``=` `'^'` `and` `p[``ord``(stC[``-``1``])] >``=` `p[``ord``(s[i])])` `                    ``or` `(s[i] ``=``=` `'^'``and` `                    ``p[``ord``(stC[``-``1``])] > p[``ord``(s[i])]))):` `              `  `                ``# Get and remove the top element` `                ``# from the character stack` `                ``t ``=` `newNode(stC[``-``1``])` `                ``stC.pop()`   `                ``# Get and remove the top element` `                ``# from the node stack` `                ``t1 ``=` `stN[``-``1``]` `                ``stN.pop()`   `                ``# Get and remove the currently top` `                ``# element from the node stack` `                ``t2 ``=` `stN[``-``1``]` `                ``stN.pop()`   `                ``# Update the tree` `                ``t.left ``=` `t2` `                ``t.right ``=` `t1`   `                ``# Push the node to the node stack` `                ``stN.append(t)`   `            ``# Push s[i] to char stack` `            ``stC.append(s[i])` `            `  `        ``elif` `(s[i] ``=``=` `')'``):` `            ``while` `(``len``(stC) !``=` `0` `and` `stC[``-``1``] !``=` `'('``):` `                ``t ``=` `newNode(stC[``-``1``])` `                ``stC.pop()` `                ``t1 ``=` `stN[``-``1``]` `                ``stN.pop()` `                ``t2 ``=` `stN[``-``1``]` `                ``stN.pop()` `                ``t.left ``=` `t2` `                ``t.right ``=` `t1` `                ``stN.append(t)` `            ``stC.pop()` `    ``t ``=` `stN[``-``1``]` `    ``return` `t`   `# Function to print the post order` `# traversal of the tree` `def` `postorder(root):` `    ``if` `(root !``=` `None``):` `        ``postorder(root.left)` `        ``postorder(root.right)` `        ``print``(root.data, end ``=` `"")`   `s ``=` `"(a^b^(c/d/e-f)^(x*y-m*n))"` `s ``=` `"("` `+` `s` `s ``+``=` `")"` `root ``=` `build(s)`   `# Function call` `postorder(root)`   `# This code is contributed by divyeshrabadiya07.`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG` `{`   `// Tree Structure` `public`  `class` `nptr` `{` `   ``public` `char` `data;` `   ``public` `nptr left, right;` `} ;`   `// Function to create new node` `static` `nptr newNode(``char` `c)` `{` `    ``nptr n = ``new` `nptr();` `    ``n.data = c;` `    ``n.left = n.right = ``null``;` `    ``return` `n;` `}`   `// Function to build Expression Tree` `static` `nptr build(String s)` `{`   `    ``// Stack to hold nodes` `    ``Stack stN = ``new` `Stack();`   `    ``// Stack to hold chars` `    ``Stack<``char``> stC = ``new` `Stack<``char``>();` `    ``nptr t, t1, t2;`   `    ``// Prioritising the operators` `    ``int` `[]p = ``new` `int``[123];` `    ``p[``'+'``] = p[``'-'``] = 1;` `    ``p[``'/'``] = p[``'*'``] = 2;` `    ``p[``'^'``] = 3;` `    ``p[``')'``] = 0;`   `    ``for` `(``int` `i = 0; i < s.Length; i++) ` `    ``{` `        ``if` `(s[i] == ``'('``)` `        ``{`   `            ``// Push '(' in char stack` `            ``stC.Push(s[i]);` `        ``}`   `        ``// Push the operands in node stack` `        ``else` `if` `(``char``.IsLetter(s[i])) ` `        ``{` `            ``t = newNode(s[i]);` `            ``stN.Push(t);` `        ``}` `        ``else` `if` `(p[s[i]] > 0) ` `        ``{` `          `  `            ``// If an operator with lower or` `            ``// same associativity appears` `            ``while` `(stC.Count != 0 && stC.Peek() != ``'('` `                ``&& ((s[i] != ``'^'` `&& p[stC.Peek()] >= p[s[i]])` `                    ``|| (s[i] == ``'^'``&& p[stC.Peek()] > p[s[i]]))) ` `            ``{`   `                ``// Get and remove the top element` `                ``// from the character stack` `                ``t = newNode(stC.Peek());` `                ``stC.Pop();`   `                ``// Get and remove the top element` `                ``// from the node stack` `                ``t1 = stN.Peek();` `                ``stN.Pop();`   `                ``// Get and remove the currently top` `                ``// element from the node stack` `                ``t2 = stN.Peek();` `                ``stN.Pop();`   `                ``// Update the tree` `                ``t.left = t2;` `                ``t.right = t1;`   `                ``// Push the node to the node stack` `                ``stN.Push(t);` `            ``}`   `            ``// Push s[i] to char stack` `            ``stC.Push(s[i]);` `        ``}` `        ``else` `if` `(s[i] == ``')'``) ` `        ``{` `            ``while` `(stC.Count != 0 && stC.Peek() != ``'('``) ` `            ``{` `                ``t = newNode(stC.Peek());` `                ``stC.Pop();` `                ``t1 = stN.Peek();` `                ``stN.Pop();` `                ``t2 = stN.Peek();` `                ``stN.Pop();` `                ``t.left = t2;` `                ``t.right = t1;` `                ``stN.Push(t);` `            ``}` `            ``stC.Pop();` `        ``}` `    ``}` `    ``t = stN.Peek();` `    ``return` `t;` `}`   `// Function to print the post order` `// traversal of the tree` `static` `void` `postorder(nptr root)` `{` `    ``if` `(root != ``null``) ` `    ``{` `        ``postorder(root.left);` `        ``postorder(root.right);` `        ``Console.Write(root.data);` `    ``}` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``String s = ``"(a^b^(c/d/e-f)^(x*y-m*n))"``;` `    ``s = ``"("` `+ s;` `    ``s += ``")"``;` `    ``nptr root = build(s);` `  `  `    ``// Function call` `    ``postorder(root);` `}` `}`   `// This code is contributed by aashish1995`

## Javascript

 ``

Output

`abcd/e/f-xy*mn*-^^^`

The time Complexity is O(n) as each character is accessed only once.
The space Complexity is O(n) as (char_stack + node_stack) <= n

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