使用中序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> v;
public void inOrder(TreeNode root){
if(root == null) {
return;
}
inOrder(root.left);
v.add(root.val);
inOrder(root.right);
}
public boolean isValidBST(TreeNode root) {
v = new ArrayList<>();
inOrder(root);
if(v.size() == 0 || v.size() == 1){
return true;
}
int tmp = v.get(0);
for(int i=1; i<v.size(); i++){
if(v.get(i) <= tmp){
return false;
}
tmp = v.get(i);
}
return true;
}
}
非递归方式实现的中序遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
stack<TreeNode*> stack;
long long inorder = (long long) INT_MIN - 1;
while(!stack.empty() || root != nullptr){
while(root != nullptr) {
stack.push(root);
root = root->left;
}
root = stack.top();
stack.pop();
// 如果中序遍历得到的节点的值小于等于前一个inorder, 说明不是二叉搜索树
if(root->val <= inorder){
return false;
}
inorder = root->val;
root = root -> right;
}
return true;
}
};
原文:https://www.cnblogs.com/eat-too-much/p/14823822.html