Challenge Description
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.
Recursive Algorithm
For checking the property of every node in search tree, we recursive traverse all the nodes.
We use a rec helper function with two more extra parameters as the left bounder and right bounder node.
class Solution {
public:
bool rec(TreeNode* root, TreeNode* min, TreeNode* max) {
if(root == NULL)
return true;
if ((min != NULL && root->val <= min->val) ||
(max != NULL && root->val >= max->val))
return false;
return rec(root->left, min, root) && rec(root->right, root, max) ;
}
bool isValidBST(TreeNode* root) {
return rec(root, NULL, NULL);
}
};
Traverse binary tree with inorder
For the property of BST, if we traverse the tree in the inorder, it will be a increasing sequence. Add a varaible name of pre to store the previous tree node.
If it is not a valid binary search tree, this property will not hold.
class Solution {
public:
TreeNode* pre;
bool in_order(TreeNode* cur) {
if(cur == NULL) return true;
if(cur->left && !in_order(cur->left))
return false;
if(pre != NULL && pre->val >= cur->val)
return false;
pre = cur;
if(cur->right && !in_order(cur->right))
return false;
return true;
}
bool isValidBST(TreeNode* root) {
pre = NULL;
return in_order(root);
}
};
We can also use a stack to implement a iteractive algorithm:
class Solution {
public:
bool isValidBST(TreeNode* root) {
TreeNode* pre = NULL;
stack<TreeNode*> stack;
TreeNode* cur = root;
while(!stack.empty() || cur != NULL) {
if(cur != NULL) {
stack.push(cur);
cur = cur->left;
} else {
cur = stack.top();
if(pre != NULL && pre->val >= cur->val)
return false;
pre = cur;
stack.pop();
cur = cur->right;
}
}
return true;
}
};
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