Validate Binary Search Tree 解答

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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 be binary search trees.

Solution 1 -- Recursive

According to the question, we can write recursive statements. Note here whole left/right subtree should be 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  */10 public class Solution {11     public boolean isValidBST(TreeNode root) {12         if (root == null)13             return true;14         if (root.left != null && !smallerThanRoot(root, root.left))15             return false;16         if (root.right != null && !greaterThanRoot(root, root.right))17             return false;18         if (isValidBST(root.left) && isValidBST(root.right))19             return true;20         return false;21     }22     23     private boolean greaterThanRoot(TreeNode root, TreeNode child) {24         if (child.val <= root.val)25             return false;26         if (child.left != null) {27             if (!greaterThanRoot(root, child.left))28                 return false;29         }30         if (child.right != null) {31             if (!greaterThanRoot(root, child.right))32                 return false;33         }34         return true;35     }36     37     private boolean smallerThanRoot(TreeNode root, TreeNode child) {38         if (child.val >= root.val)39             return false;40         if (child.left != null) {41             if (!smallerThanRoot(root, child.left))42                 return false;43         }44         if (child.right != null) {45             if (!smallerThanRoot(root, child.right))46                 return false;47         }48         return true;49     }50 }

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  */10 public class Solution {11     public boolean isValidBST(TreeNode root) {12         // This problem can be looked as inorder traversal problem13         // Inorder traversal of BST is an ascending array14         List<Integer> inOrderResult = new ArrayList<Integer>();15         Stack<TreeNode> stack = new Stack<TreeNode>();16         TreeNode tmp = root;17         while (tmp != null || !stack.empty()) {18             if (tmp != null) {19                 stack.push(tmp);20                 tmp = tmp.left;21             } else {22                 TreeNode current = stack.pop();23                 inOrderResult.add(current.val);24                 tmp = current.right;25             }26         }27         // Traverse list28         if (inOrderResult.size() < 1)29             return true;30         int max = inOrderResult.get(0);31         for (int i = 1; i < inOrderResult.size(); i++) {32             if (inOrderResult.get(i) > max)33                 max = inOrderResult.get(i);34             else35                 return false;36         }37         return true;38     }39 }

 

Validate Binary Search Tree 解答