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.
Example:
Input:
2
/ \
1 3
Output: true
解析
判断二叉树是否为BST,递归判断左右子树是否为BST,并判断左子树最大值小于根节点的值,右子树最小值大于根节点的值。,可以使用递归和中序遍历两种方法来解决。
解法1:递归
递归判断左右子树和左右最大的值,但是需要考虑INT的最大最小值情况,所以使用LONG。
class Solution {
public:
bool isValidBST(TreeNode* root) {
if(!root)
return true;
if(!root->left && !root->right)
return true;
if(root->left || root->right){
long left_max = LONG_MIN;
long right_min = LONG_MAX;
TreeNode* p = root->left;
while(p){
left_max = max(left_max, long(p->val));
p = p->right;
}
p = root->right;
while(p){
right_min = min(right_min, long(p->val));
p = p->left;
}
return left_max < root->val && right_min > root->val && isValidBST(root->left) && isValidBST(root->right) ;
}
return false;
}
};
还可以构建新的递归函数来解决。
class Solution {
public:
bool isValidBST(TreeNode* root) {
return isValidBST(root, LONG_MIN, LONG_MAX);
}
bool isValidBST(TreeNode* root, long left_max, long right_min){
if(!root)
return true;
if(left_max >= root->val || right_min <= root->val)
return false;
return isValidBST(root->left, left_max, root->val) && isValidBST(root->right, root->val, right_min);
}
};
解法2:中序遍历
可以选择得到完整的中序遍历判断是否有序,也可以在得到中序遍历的过程中判断是否有序,对于中序遍历又有好几种方法,具体可以参考Binary Tree Inorder Traversal。
参考
http://www.cnblogs.com/grandyang/p/4298435.html
https://blog.youkuaiyun.com/Peng_maple/article/details/82155101