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.
2 / \ 1 3 Input: [2,1,3] Output: true
5 / \ 1 4 / \ 3 6 Input: [5,1,4,null,null,3,6] Output: false Explanation: The root node's value is 5 but its right child's value is 4.
通过递归,判断当前节点的左节点是否小于当前节点,右节点是否大于当前节点。这里需要注意的是子树所有节点都要满足条件。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return helper(root, null, null);
}
public boolean helper(TreeNode node, TreeNode min, TreeNode max) {
if (node == null) return true;
if (min != null && node.val <= min.val) return false;
if (max != null && node.val >= max.val) return false;
if (!helper(node.left, min, node)) return false;
if (!helper(node.right, node, max)) return false;
return true;
}
}
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
return helper(root, NULL, NULL);
}
bool helper(TreeNode* node, TreeNode* min, TreeNode* max) {
if (node == NULL) return true;
if (max != NULL && node->val >= max->val) return false;
if (min != NULL && node->val <= min->val) return false;
if (!helper(node->left, min, node)) return false;
if (!helper(node->right, node, max)) return false;
return true;
}
};