07
08/2015
[LeetCode] Validate Binary Search Tree
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.
解题思路：
（1）空二叉树是BST
（2）若根节点R值大于左子树的最大值，且小于右子树最小值，并且左右子树都是BST，那么以R为根的树也是BST
因此代码如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
if(root == NULL){
return true;
}
TreeNode* leftMaxNode = getMaxNode(root->left);
TreeNode* rightMinNode = getMinNode(root->right);
if(leftMaxNode!=NULL && leftMaxNode->val >= root->val){
return false;
}
if(rightMinNode!=NULL && rightMinNode->val <= root->val){
return false;
}
return isValidBST(root->left) && isValidBST(root->right);
}
TreeNode* getMaxNode(TreeNode* root){
if(root == NULL){
return NULL;
}
TreeNode* maxNode = root;
TreeNode* leftMaxNode = getMaxNode(root->left);
TreeNode* rightMaxNode = getMaxNode(root->right);
if(leftMaxNode!=NULL && leftMaxNode->val > maxNode->val){
maxNode = leftMaxNode;
}
if(rightMaxNode!=NULL && rightMaxNode->val > maxNode->val){
maxNode = rightMaxNode;
}
return maxNode;
}
TreeNode* getMinNode(TreeNode* root){
if(root == NULL){
return NULL;
}
TreeNode* minNode = root;
TreeNode* leftMinNode = getMinNode(root->left);
TreeNode* rightMinNode = getMinNode(root->right);
if(leftMinNode!=NULL && leftMinNode->val < minNode->val){
minNode = leftMinNode;
}
if(rightMinNode!=NULL && rightMinNode->val < minNode->val){
minNode = rightMinNode;
}
return minNode;
}
};
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