我们说我有以下后缀表达式:5372 - * -
我想从这个表达式创建一个二叉树。我的算法是:如果我的char是数字,如果它是一个操作符从堆栈中弹出两个元素并将它们作为运算符的子元素,则将其放入堆栈。然后将操作员推入堆栈。这似乎有效,但我无法连接我创建的小树。
我目前的代码是:
public void myInsert(char ch, Stack s) {
if (Character.isDigit(ch)) // initial cond.
s.push(ch);
else {
TreeNode tParent = new TreeNode(ch);
TreeNode t = new TreeNode(s.pop());
TreeNode t2 = new TreeNode(s.pop());
tParent.right = t;
tParent.left = t2;
s.push(ch);
System.out.println("par" + tParent.ch);
System.out.println("cright" + tParent.right.ch);
System.out.println("cleft" + tParent.left.ch);
}
}
我的测试班:
Stack stree = new Stack();
BST b = new BST();
String str = "5-3*(7-2)";
String postfix = b.convertToPosFix(str);
System.out.println(postfix);
for (char ch : postfix.toCharArray()) {
b.myInsert(ch, stree);
}
我的输出是:
par-
cright2
cleft7
par*
cright-
cleft3
par-
cright*
cleft5
答案 0 :(得分:1)
使用Stack
个TreeNode
,而不是Stack
个字符。您必须首先合并TreeNode
s而不是char
s:
public void myInsert(char ch, Stack<TreeNode> s) {
if (Character.isDigit(ch)) {
// leaf (literal)
s.push(new TreeNode(ch));
} else {
// operator node
TreeNode tParent = new TreeNode(ch);
// add operands
tParent.right = s.pop();
tParent.left = s.pop();
// push result to operand stack
s.push(tParent);
}
}
public class TreeNode {
public TreeNode right = null;
public TreeNode left = null;
public final char ch;
TreeNode(char ch) {
this.ch = ch;
}
@Override
public String toString() {
return (right == null && left == null) ? Character.toString(ch) : "(" + left.toString()+ ch + right.toString() + ")";
}
}
public static TreeNode postfixToTree(String s) {
Stack<TreeNode> stree = new Stack<>();
BST b = new BST();
for (char ch : s.toCharArray()) {
b.myInsert(ch, stree);
}
return stree.pop();
}
public static void main(String[] args) {
System.out.println(postfixToTree("5372-*-"));
System.out.println(postfixToTree("512+4*+3−"));
System.out.println(postfixToTree("51*24*+"));
}
这将打印
(5-(3*(7-2)))
((5+((1+2)*4))−3)
((5*1)+(2*4))