Expression Tree Build

The structure of Expression Tree is a binary tree to evaluate certain expressions. All leaves of the Expression Tree have an number string value. All non-leaves of the Expression Tree have an operator string value.

Now, given an expression array, build the expression tree of this expression, return the root of this expression tree.

Have you met this question in a real interview?

Example

For the expression (2*6-(23+7)/(1+2)) (which can be represented by [“2” “*” “6” “-” “(” “23” “+” “7” “)” “/” “(” “1” “+” “2” “)”]). The expression tree will be like

                 [ - ]
             /          \
        [ * ]              [ / ]
      /     \           /         \
    [ 2 ]  [ 6 ]      [ + ]        [ + ]
                     /    \       /      \
                   [ 23 ][ 7 ] [ 1 ]   [ 2 ] .

After building the tree, you just need to return root node [-].

Clarification

See wiki: Expression Tree

Tags Expand

Related Problems Expand

Use two stacks, operand and operator.

/**
 * Definition of ExpressionTreeNode:
 * class ExpressionTreeNode {
 * public:
 *     string symbol;
 *     ExpressionTreeNode *left, *right;
 *     ExpressionTreeNode(string symbol) {
 *         this->symbol = symbol;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param expression: A string array
     * @return: The root of expression tree
     */
    ExpressionTreeNode* build(vector<string> &expression) {
        // write your code here
        stack<ExpressionTreeNode *> operators, operands;
        for(int i = 0; i < expression.size(); ++i){
            auto exp = expression[i];
            ExpressionTreeNode * p = new ExpressionTreeNode(exp);
            if(exp[0] <= '9' || exp[0] >= '0'){
                //digit
                operands.push(p);
            }
            else{
                //operator
                if(operators.empty()){
                    operators.push(p);
                }else if(exp[0] == '('){
                    operators.push(p);
                }else if(exp[0] == ')'){
                    while(operators.top() != '('){
                        eval(operators, operands);
                    }
                    operators.pop();//pop ')'
                }else if(getPriority(p) > getPriority(operators.top())){
                    operators.push(p);
                }else{
                    //evaluate 
                    eval(operators, operands);
                    delete p;
                    i--;
                }
            }
        }
        while(!operators.empty()){
            eval(operators, operands);
        }
        return operands.top();
    }
    void eval(stack<ExpressionTreeNode *>& operators, stack<ExpressionTreeNode *>& operands){
        ExpressionTreeNode * oper = operators.top();
        operators.pop();
        ExpressionTreeNode * operand1 = operands.top();
        operands.pop();
        ExpressionTreeNode * operand2 = operands.top();
        operands.pop();
        oper->right = operand1;
        oper->left = operand2;
        operands.push(oper);
    }
    int getPriority(ExpressionTreeNode* a){
        char c = a->symbol[0];
        if(c == '+' || c == '-'){
            return 1;
        }
        else if(c == '*' || c == '/'){
            return 2;
        }
    }
};