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.
思路：中序遍历。当前值要比之前的小。
bool isValidBST(TreeNode* root) {
TreeNode * pPre = NULL;
TreeNode * pCur = root;
vector<TreeNode *> v;
while(!v.empty() || NULL != pCur)
{
if(NULL != pCur)
{
v.push_back(pCur);
pCur = pCur->left;
}
else
{
if(pPre != NULL && v.back()->val <= pPre->val)
return false;
pPre = v.back();
v.pop_back();
pCur = pPre->right;
}
}
return true;
}
大神递归版：注意，每次左子树的值范围在最小值和根值之间，右子树的范围在根植和最大值之间。
public class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (root == null) return true;
if (root.val >= maxVal || root.val <= minVal) return false;
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
}