用二叉树后序遍历计算中缀表达式的值

如果用一颗二叉树表示加减乘除及0-9的数字构成的数学表达式，对二叉树进行后序遍历得到的就是后缀表达式。后缀表达式可以通过堆栈直接计算其值。

 

如果输入的是中缀表达式，下面的方法toBinaryTree()将它转化为二叉树。

 

代码：

import java.util.ArrayList; import java.util.List; import java.util.Stack; /** * * @author ljs * 2011-05-24 * * 把中缀表达式转换成一棵二叉树，然后通过后序遍历计算表达式的值 * 1. unary operator like -3 is not supported * 2. only support + - * / and digits 0-9 * */ public class InfixExpression { public class Node{ char c; Node left; Node right; public Node(char c){ this.c = c; } } private String infixExpr; private Node binTree; public InfixExpression(String infixExpr){ this.infixExpr = infixExpr; } public Node toBinaryTree() throws Exception{ binTree = toBinaryTree(this.infixExpr); return binTree; } public double evalUsingPostfix(){ List<Character> postfixSymbols = new ArrayList<Character>(); postOrderTraverse(this.binTree,postfixSymbols); Stack<Double> stack = new Stack<Double>(); for(char c:postfixSymbols){ switch(c){ case '+':{ double op2 = stack.pop(); double op1 = stack.pop(); double result = op1 + op2; stack.push(result); } break; case '-':{ double op2 = stack.pop(); double op1 = stack.pop(); double result = op1 - op2; stack.push(result); } break; case '*':{ double op2 = stack.pop(); double op1 = stack.pop(); double result = op1 * op2; stack.push(result); } break; case '/':{ double op2 = stack.pop(); double op1 = stack.pop(); double result = op1 / op2; stack.push(result); } break; default: double val = c-'0'; stack.push(val); break; } } return stack.pop(); } private void postOrderTraverse(Node root,List<Character> postfixSymbols){ if(root == null) return; //traverse using postorder postOrderTraverse(root.left,postfixSymbols); postOrderTraverse(root.right,postfixSymbols); postfixSymbols.add(root.c); } private Node toBinaryTree(String subexpr) throws Exception{ subexpr = subexpr.trim(); int len = subexpr.length(); boolean checkRight = false; Node rootNode = null; Node leftNode = null; int i = 0; Node tree=null; while(i<len) { char c = subexpr.charAt(i); if(c>='0' && c<='9'){ i++; if(checkRight){ checkRight = false; if(rootNode == null) throw new Exception("input is not a valid infix expr."); rootNode.right = new Node(c); //for expr. 2+5+3 when c='5' leftNode = rootNode; }else{ leftNode = new Node(c); } }else if(c=='('){ Stack<Character> stack = new Stack<Character>(); StringBuffer sb = new StringBuffer(); int p = i; stack.push('('); while(p<len){ p++; char symbol = subexpr.charAt(p); if(symbol=='('){ sb.append(symbol); stack.push(symbol); }else if(symbol==')'){ if(stack.isEmpty()){ throw new Exception("input is not a valid infix expr."); }else{ stack.pop(); } if(stack.isEmpty()) { //the last symbol ) is removed break; }else{ sb.append(symbol); } }else{ sb.append(symbol); } } if(!stack.isEmpty()) throw new Exception("input is not a valid infix expr."); String subtreeExpr=sb.toString(); Node subtree = toBinaryTree(subtreeExpr); i = ++p; if(checkRight){ checkRight = false; if(rootNode == null) throw new Exception("input is not a valid infix expr."); rootNode.right = subtree; //for expr. 1+(2+5)+3 when subtree = (2+5) leftNode = rootNode; }else{ leftNode = subtree; } }else if(c=='+' || c=='-' || c=='*' || c=='/'){ i++; if(leftNode == null){ throw new Exception("input is not a valid infix expr."); }else{ Node node = new Node(c); node.left = leftNode; rootNode = node; checkRight = true; } }else if(c==' '){ i++; //ignore blank char }else{ throw new Exception("input is not a valid infix expr."); } } if(i<len) throw new Exception("input is not a valid infix expr."); tree = leftNode; return tree; } public static void main(String[] args) throws Exception { String infixExpr ="8-((3+5)*(5-(6/2)))"; //String infixExpr ="(2+2)*(2+2)"; //String infixExpr ="((2 + 2 ) * 3 + 3 )/2"; //String infixExpr ="(((2-6)+((7*2)*3))-5)"; InfixExpression infix = new InfixExpression(infixExpr); Node binTree = infix.toBinaryTree(); double result = infix.evalUsingPostfix(); System.out.format("%s = %.2f%n",infixExpr,result); } }

 

 

测试：

8-((3+5)*(5-(6/2))) = -8.00