Validate Binary Search Tree Solution

Question

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 is binary search trees.

Solution 1--Recursive

According to the question, we can write recursive statements. Note here whole left/right subtree should is smaller/greater than the root.

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      Public BooleanIsvalidbst (TreeNode root) { A         if(Root = =NULL) -             return true; -         if(Root.left! =NULL&&!smallerthanroot (Root, Root.left)) the             return false; -         if(Root.right! =NULL&&!greaterthanroot (Root, root.right)) -             return false; -         if(Isvalidbst (Root.left) &&Isvalidbst (root.right)) +             return true; -         return false; +     } A      at     Private Booleangreaterthanroot (TreeNode root, TreeNode child) { -         if(Child.val <=root.val) -             return false; -         if(Child.left! =NULL) { -             if(!greaterthanroot (Root, Child.left)) -                 return false; in         } -         if(Child.right! =NULL) { to             if(!greaterthanroot (Root, child.right)) +                 return false; -         } the         return true; *     } $     Panax Notoginseng     Private Booleansmallerthanroot (TreeNode root, TreeNode child) { -         if(Child.val >=root.val) the             return false; +         if(Child.left! =NULL) { A             if(!smallerthanroot (Root, Child.left)) the                 return false; +         } -         if(Child.right! =NULL) { $             if(!smallerthanroot (Root, child.right)) $                 return false; -         } -         return true; the     } -}

Solution 2--inorder traversal

Inorder traversal of BST is an ascending array. Java Stack

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      Public BooleanIsvalidbst (TreeNode root) { A         //This problem can looked as inorder traversal problem -         //inorder Traversal of BST is an ascending array -List<integer> Inorderresult =NewArraylist<integer>(); thestack<treenode> stack =NewStack<treenode>(); -TreeNode tmp =Root; -          while(TMP! =NULL|| !Stack.empty ()) { -             if(TMP! =NULL) { + Stack.push (TMP); -TMP =Tmp.left; +}Else { ATreeNode current =Stack.pop (); at Inorderresult.add (current.val); -TMP =Current.right; -             } -         } -         //Traverse List -         if(Inorderresult.size () < 1) in             return true; -         intmax = Inorderresult.get (0); to          for(inti = 1; I < inorderresult.size (); i++) { +             if(Inorderresult.get (i) >max) -Max =Inorderresult.get (i); the             Else *                 return false; $         }Panax Notoginseng         return true; -     } the}

Validate Binary Search Tree Solution