Java Expression Binary Tree / Equation Binary Tree

populate method for prefix and postfix from the provided code. Method headers included in the file The populate methods should not convert to the other type before populating the tree nodes. Add Big-O notation to all 3 populate methods.

========================================provided code=============================

public class EquationBinaryTree {

private Node root;

public EquationBinaryTree()

{

root = null;

}

//(left parent right)

//(->left->parent->right->)

public void printInfix()

{

printInfixHelper(root);

System.out.println();

}

private void printInfixHelper(Node n)//O(N) - visit each node only once

{

if(n != null && n.left != null)

{

System.out.print("(");

printInfixHelper(n.left);//left side

System.out.print(n.value);//middle item//parent

printInfixHelper(n.right);//right side

System.out.print(")");

}

else if(n != null)

{

System.out.print(n.value);//middle item//parent

}

}

//left -> right -> parent

public void printPostfix()

{

printPostfixHelper(root);

System.out.println();

}

private void printPostfixHelper(Node n)

{

if(n != null && n.left != null)

{

printPostfixHelper(n.left);//left side

printPostfixHelper(n.right);//right side

System.out.print(n.value);//middle item//parent

}

else if(n != null)

{

System.out.print(n.value);//middle item//parent

}

}

//parent -> left -> right

public void printPrefix()

{

printPrefixHelper(root);

System.out.println();

}

private void printPrefixHelper(Node n)

{

if(n != null && n.left != null)

{

System.out.print(n.value);//middle item//parent

printPrefixHelper(n.left);//left side

printPrefixHelper(n.right);//right side

}

else if(n != null)

{

System.out.print(n.value);//middle item//parent

}

}

//(a+b)

//(a+(b*c))

//((a*b)+c)

public void populateFromInfix(String equation)

{

root = populateFromInfixHelper(equation);

}

public Node populateFromInfixHelper(String equation)

{

if(equation.length() == 1)

{

return new Node(equation);//math operand

}

//System.out.println(equation);

String temp = equation.substring(1, equation.length()-1);//remove wrapper paren

//begin search for middle of equation

int parenCount = 0;

int mid = 0;

for(int i = 0; i < temp.length(); i++)

{

if(temp.charAt(i) == '(')

parenCount++;

if(temp.charAt(i) == ')')

parenCount--;

if(parenCount == 0)

{

mid = i+1;

break;

}

}

//middle

Node n = new Node(""+temp.charAt(mid));

//first half

n.left = populateFromInfixHelper(temp.substring(0, mid));

//System.out.println(temp.substring(0, mid));

//second half

n.right = populateFromInfixHelper(temp.substring(mid+1));

//System.out.println(temp.substring(mid+1));

return n;

}

//ab+ = (a+b)

public void populateFromPostfix(String equation)

{

//complete this section for #5 from Assignment 4

//does not require recursion to solve

root = null;

}

//+ab = (a+b)

public void populateFromPrefix(String equation)

{

//complete this section for #5 from Assignment 4

//does not require recursion to solve

root = null;

}

private class Node

{

String value;

Node left, right;

public Node(String v)

{

value = v;

left = null;

right = null;

}

}

}