package problems.medium;
import problems.utils.TreeNode;
import java.util.ArrayList;
import java.util.List;
/**
* Created by sherxon on 1/5/17.
*/
public class ValidateBinarySearchTree {
/**
* Solution 1:
* solution one building valid bst from preorder traversal of input tree and check if they are the same
*/
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
List<Integer> prOrder = new ArrayList<>(); // hashset is more appropriate actually
dfs(root, prOrder);
TreeNode x = null;
for (Integer i : prOrder)
x = insert(x, i);
return isTheSameTree(x, root);
}
boolean isTheSameTree(TreeNode x, TreeNode root) {
if (x == null || root == null) return x == root;
return x.val == root.val && isTheSameTree(x.left, root.left) && isTheSameTree(x.right, root.right);
}
TreeNode insert(TreeNode x, int val) {
if (x == null) return new TreeNode(val);
if (x.val < val)
x.right = insert(x.right, val);
else if (x.val > val)
x.left = insert(x.left, val);
return x;
}
void dfs(TreeNode x, List<Integer> list) {
if (x == null) return;
list.add(x.val);
dfs(x.left, list);
dfs(x.right, list);
}
/**
* Solution 2
* Basically what I am doing is recursively iterating over the tree while defining interval <minVal, maxVal> for each node which value must fit in.
* */
public boolean isValidBST2(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (root == null) return true;
if (root.val >= maxVal || root.val <= minVal) return false;
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
}