用二叉樹後序遍歷計算中綴表達式的值

http://blog.csdn.net/ljsspace/article/details/6442921

 

如果用一顆二叉樹表示加減乘除及0-9的數字構成的數學表達式，對二叉樹進行後序遍歷得到的就是後綴表達式。後綴表達式可以通過堆棧直接計算其值。

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
/**
 * 
 * @author ljs
 * 2011-05-24
 * 
 * 把中綴表達式轉換成一棵二叉樹，然後通過後序遍歷計算表達式的值
 * 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);
	}
}

 

如果輸入的是中綴表達式，下面的方法toBinaryTree()將它轉化爲二叉樹。