Evaluation of Expression Tree
Given a simple expression tree, consisting of basic binary operators i.e., + , – ,* and / and some integers, evaluate the expression tree.
Examples:
Input : Root node of the below treeOutput : 100 Input : Root node of the below tree
Output : 110
As all the operators in the tree are binary hence each node will have either 0 or 2 children. As it can be inferred from the examples above , the integer values would appear at the leaf nodes , while the interior nodes represent the operators.
To evaluate the syntax tree , a recursive approach can be followed .
Algorithm :
Let t be the syntax tree
If t is not null then
If t.info is operand then
Return t.info
Else
A = solve(t.left)
B = solve(t.right)
return A operator B
where operator is the info contained in t
The time complexity would be O(n), as each node is visited once. Below is a C++ program for the same:
C/C++
// C++ program to evaluate an expression tree #include <bits/stdc++.h> using namespace std; // Class to represent the nodes of syntax tree class node { public: string info; node *left = NULL, *right = NULL; node(string x) { info = x; } }; // Utility function to return the integer value // of a given string int toInt(string s) { int num = 0; // Check if the integral value is // negative or not // If it is not negative, generate the number // normally if(s[0]!='-') for (int i=0; i<s.length(); i++) num = num*10 + (int(s[i])-48); // If it is negative, calculate the +ve number // first ignoring the sign and invert the // sign at the end else for (int i=1; i<s.length(); i++) { num = num*10 + (int(s[i])-48); num = num*-1; } return num; } // This function receives a node of the syntax tree // and recursively evaluates it int eval(node* root) { // empty tree if (!root) return 0; // leaf node i.e, an integer if (!root->left && !root->right) return toInt(root->info); // Evaluate left subtree int l_val = eval(root->left); // Evaluate right subtree int r_val = eval(root->right); // Check which operator to apply if (root->info=="+") return l_val+r_val; if (root->info=="-") return l_val-r_val; if (root->info=="*") return l_val*r_val; return l_val/r_val; } //driver function to check the above program int main() { // create a syntax tree node *root = new node("+"); root->left = new node("*"); root->left->left = new node("5"); root->left->right = new node("-4"); root->right = new node("-"); root->right->left = new node("100"); root->right->right = new node("20"); cout << eval(root) << endl; delete(root); root = new node("+"); root->left = new node("*"); root->left->left = new node("5"); root->left->right = new node("4"); root->right = new node("-"); root->right->left = new node("100"); root->right->right = new node("/"); root->right->right->left = new node("20"); root->right->right->right = new node("2"); cout << eval(root); return 0; } |
Python
# Python program to evaluate expression tree # Class to represent the nodes of syntax treeclass node: def __init__(self, value): self.left = None self.data = value self.right = None # This function receives a node of the syntax tree# and recursively evaluate itdef evaluateExpressionTree(root): # empty tree if root is None: return 0 # leaf node if root.left is None and root.right is None: return int(root.data) # evaluate left tree left_sum = evaluateExpressionTree(root.left) # evaluate right tree right_sum = evaluateExpressionTree(root.right) # check which operation to apply if root.data == '+': return left_sum + right_sum elif root.data == '-': return left_sum - right_sum elif root.data == '*': return left_sum * right_sum else: return left_sum / right_sum # Driver function to test above problemif __name__=='__main__': # creating a sample tree root = node('+') root.left = node('*') root.left.left = node('5') root.left.right = node('4') root.right = node('-') root.right.left = node('100') root.right.right = node('20') print evaluateExpressionTree(root) root = None #creating a sample tree root = node('+') root.left = node('*') root.left.left = node('5') root.left.right = node('4') root.right = node('-') root.right.left = node('100') root.right.right = node('/') root.right.right.left = node('20') root.right.right.right = node('2') print evaluateExpressionTree(root) # This code is contributed by Harshit Sidhwa |
Output:
60 110
This article is contributed by Ashutosh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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