Given a simple expression tree, consisting of basic binary operators i.e., + , – ,* and / and some integers, evaluate the expression tree.
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:
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- Stack | Set 4 (Evaluation of Postfix Expression)
- Building Expression tree from Prefix Expression
- Expression Tree
- Convert Ternary Expression to a Binary Tree
- Convert ternary expression to Binary Tree using Stack
- Program to convert Infix notation to Expression Tree
- Complexity of different operations in Binary tree, Binary Search Tree and AVL tree
- Maximum sub-tree sum in a Binary Tree such that the sub-tree is also a BST
- Convert a Generic Tree(N-array Tree) to Binary Tree
- Minimum number of bracket reversals needed to make an expression balanced
- Check for Balanced Brackets in an expression (well-formedness) using Stack
- Convert a Binary Tree into its Mirror Tree
- Convert an arbitrary Binary Tree to a tree that holds Children Sum Property
- Check if a binary tree is subtree of another binary tree | Set 1
- Convert a given tree to its Sum Tree
- Binary Tree to Binary Search Tree Conversion
- Check if a given Binary Tree is height balanced like a Red-Black Tree
- Check if a binary tree is subtree of another binary tree | Set 2
- Convert a Binary Tree to Threaded binary tree | Set 1 (Using Queue)
- Check whether a binary tree is a full binary tree or not