How to convert a Ternary expression to a Binary tree structure?
This one would be without using parent node. But using stack.
public NodeC convertTtoBT (char[] values) {
char xx = values[0];
NodeC n = new NodeC(xx);
Stack<NodeC> a = new Stack<NodeC>();
for (int i = 1; i < values.length; i += 2) {
if (values[i] == '?') {
n.left = new NodeC (values[i + 1]);
a.add(n);
n = n.left;
}
else if (values[i] == ':') {
n = a.pop();
while (n.right != null) {
n = a.pop();
}
n.right = new NodeC (values[i + 1]);
a.add(n);
n = n.right;
}
}
return n;
}
Walk through the expression from right to left, pushing any letters as nodes to the stack. If you see '?', then instead of pushing the next letter, take it as the root, pop two last nodes from the stack as left and right root's children, and push the root back into the stack.
public TreeNode ternaryToTree(char[] exp) {
LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
for (int i = exp.length-1; i >= 0; i--) {
if (exp[i] == ':') continue;
if (exp[i] == '?') {
TreeNode node = new TreeNode(exp[--i]);
node.left = stack.pop();
node.right = stack.pop();
stack.push(node);
} else {
stack.push(new TreeNode(exp[i]));
}
}
return stack.isEmpty() ? null : stack.pop();
}
Below is my solution which has been tested thoroughly.
class TreeNode {
char c;
TreeNode left;
TreeNode right;
TreeNode(char c) {
this.c = c;
left = null;
right = null;
}
}
public TreeNode tenaryToTree(String s) {
if (s.length() == 0) {
return null;
}
TreeNode root = new TreeNode(s.charAt(0));
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) == '?') {
TreeNode node = stack.peek();
node.left = new TreeNode(s.charAt(i + 1));
stack.push(node.left);
} else if (s.charAt(i) == ':') {
stack.pop();
TreeNode node = stack.pop();
node.right = new TreeNode(s.charAt(i + 1));
stack.push(node.right);
}
}
return root;
}