假设我有以下后缀表达式: 5372-*-
我想从这个表达式中创建一个二叉树。我的algoritm是:如果我的char是数字,那么将它放入一个堆栈中,如果它是一个操作符,则从堆栈中弹出两个元素,并使它们成为运算符的子类。然后将操作符推到堆栈中。这看起来很管用,但我无法将我创造的小树连接起来。
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);
}
}我的考试课:
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);
}我的产出是:
par-
cright2
cleft7
par*
cright-
cleft3
par-
cright*
cleft5发布于 2015-03-21 21:40:31
使用Stack of TreeNodes,而不是chars的Stack。毕竟,您必须组合TreeNodes,而不是chars:
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);
}
}TreeNode
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() + ")";
}
}main
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*+"));
}这个会打印出来
(5-(3*(7-2)))
((5+((1+2)*4))−3)
((5*1)+(2*4))https://stackoverflow.com/questions/29187998
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