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.
解题思路:
先中序遍历二叉树,将二叉树的每个节点的值保存在一个数组中,然后判断该数组是不是严格递增,如果是,则是一棵二分查找树,否则,不是二分查找树。
class Solution {
public:
bool isValidBST(TreeNode* root) {
if (!root)
return true;
vector<int> nodeVals;
midRootTravel(root, nodeVals);
for (int i = 0; i != nodeVals.size()-1; i++){
if (nodeVals[i] >= nodeVals[i + 1])
return false;
}
return true;
}
void midRootTravel(TreeNode* root, vector<int>& nodeVals){
if (!root)
return;
midRootTravel(root->left, nodeVals);
nodeVals.push_back(root->val);
midRootTravel(root->right, nodeVals);
}
};