2.25 Validate Binary Search Tree

Description

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search trees. Example 1: 2 / \ 1 3 Binary tree [2,1,3], return true. Example 2: 1 / \ 2 3 Binary tree [1,2,3], return false.

Method

one : recursive use min and max value to check the val

two : iteration inorder traversal if node.val <= pre.val it false

Time and Space Complexity

both O(n)

Code

one :

 public boolean isValidBST(TreeNode root) {
      if (root == null){
          return true;
      }
      return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}

private boolean helper(TreeNode root, long min, long max){
        if (root == null){
            return true;
        }
        if (root.val >= max || root.val <= min){
            return false;
        }
        return helper(root.left, min, root.val) && helper(root.right, root.val, max);
}

two:

   public boolean isValidBST(TreeNode root){
          if (root == null){
          return true;
          } 
          Stack<TreeNode> stack = new Stack<TreeNode>();
          TreeNode pre = null;
          TreeNode cur = root;
          while (!stack.isEmpty() || cur != null){
                 if (cur != null){
                     stack.push(cur);
                     cur = cur.left;
                 } else {
                      TreeNode p = stack.pop();
                      if (pre != null && p.val <= pre.val){
                          return false;
                      }
                      pre = p;
                      cur = p.right;
                 }
          }
          return true;

   }