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.
Perform an in-order tree traversal, save all values in an array, and later check if it is sorted
There is room for space optimization in solution below
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void rec (TreeNode *root, vector<int> &v) {
if (root->left != NULL) {
rec(root->left,v);
}
v.push_back(root->val);
if (root->right != NULL) {
rec(root->right,v);
}
}
bool isValidBST(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root == NULL) return true;
vector<int> v;
rec(root,v);
if (v.size() < 2) {
return true;
}
// check if it's already sorted
for (int i = 1; i < v.size(); i++) {
if (v[i-1] >= v[i]) {
return false;
}
}
return true;
}
};
Update: Jan-16-2014http://leetcode.com/2010/09/determine-if-binary-tree-is-binary.html
Solution#2, in space.
Note int &pre in arguments
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool inorder (TreeNode *root, int &pre) {
if (root == NULL) {
return true;
}
if (inorder(root->left,pre)) {
if (pre >= root->val) {
return false;
}
pre = root->val;
return inorder(root->right,pre);
}
return false;
}
bool isValidBST(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root == NULL) return true;
int pre = INT_MIN;
return inorder(root,pre);
}
};
Update: Jan-15-2015
to handle INT_MIN case
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool inorder(TreeNode *root, TreeNode *&pre) {
if (root == NULL) {
return true;
}
if (inorder(root->left,pre)) {
if (pre != NULL && pre->val >= root->val) {
return false;
}
pre = root;
return inorder(root->right,pre);
}
return false;
}
bool isValidBST(TreeNode *root) {
TreeNode *pre = NULL;
return inorder(root,pre);
}
};
Java, updated Jun-24th-2018
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private TreeNode pre = null;
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
if (!isValidBST(root.left)) {
return false;
}
if (pre == null) pre = root;
else if (pre.val >= root.val) {
return false;
}
pre = root;
return isValidBST(root.right);
}
}
当有node的值等于MIN_VALUE或MAX_VALUE, 这算法就搞不定了。所有还是别用了
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return helper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
private boolean helper(TreeNode root, int min, int max) {
if (root == null) return true;
if (root.val < max && root.val > min) {
return helper(root.left, min, root.val)
&& helper(root.right, root.val, max);
}
return false;
}
}
