从后缀表达式创建二叉树
Create a Binary Tree from postfix expression
假设我有以下后缀表达式:5372-*-
我想根据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,如果它是一个运算符,则将它放入堆栈中,从堆栈中弹出两个元素,并使它们成为运算符的子元素。然后将运算符压入栈中。这似乎是可行的,但我无法连接我创建的小树。
我当前的代码是:
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class:
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
使用 Stack
个 TreeNode
个字符,而不是 Stack
个字符。毕竟你必须结合 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);
}
}
树节点
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() + ")";
}
}
主要
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))
假设我有以下后缀表达式:5372-*-
我想根据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,如果它是一个运算符,则将它放入堆栈中,从堆栈中弹出两个元素,并使它们成为运算符的子元素。然后将运算符压入栈中。这似乎是可行的,但我无法连接我创建的小树。
我当前的代码是:
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class:
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
使用 Stack
个 TreeNode
个字符,而不是 Stack
个字符。毕竟你必须结合 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);
}
}
树节点
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() + ")";
}
}
主要
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))