Validate Binary Search Tree

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.

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

思路：根据题目要求，实现三个方面判断，左子树、右子树以及左子树小于根节点，右子树大于根节点

java代码：

1. public boolean isValidBST(TreeNode root) {
2. if(root==null) return true;
3. if(root.left == null && root.right == null) return true;
4. boolean flag = isValidBST(root.left) && isValidBST(root.right);
5. if(flag==false) return false;
6. TreeNode left = root.left;
7. if(left!=null) {
8. while(left.right!=null) {
9. left = left.right;
10. }
11. flag &= (left.val<root.val);
12. }
13. TreeNode right = root.right;
14. if(right!=null) {
15. while(right.left!=null) {
16. right = right.left;
17. }
18. flag &= (right.val > root.val);
19. }
20. return flag;
21. }

C++代码：

1. bool isValidBST(TreeNode *root) {
2. bool lflag=false;
3. bool rflag=false;
4. if(root==NULL || (root->left==NULL && root->right==NULL)) return true;
5. if(isValidBST(root->left)) {
6. TreeNode *p=root->left;
7. while(p!=NULL && p->right!=NULL) {
8. p=p->right;
9. }
10. if(p==NULL || p->val<root->val) lflag=true;
11. }
12. if(isValidBST(root->right)) {
13. TreeNode *p=root->right;
14. while(p!=NULL && p->left!=NULL) {
15. p=p->left;
16. }
17. if(p==NULL || p->val>root->val) rflag=true;
18. }
19. if(lflag&&rflag) return true;
20. return false;
21. }