https://leetcode.com/problems/validate-binary-search-tree/

/**
 * 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) {
		if (root == nullptr || (root->left == nullptr && root->right == nullptr)) {
			return true;
		}
		if (root->left != nullptr && root->right != nullptr) {
			if (root->left->val >= root->val || root->right->val <= root->val) {
				return false;
			}
			return isValidBST(root->left) && isValidBST(root->right);
		}
		if (root->left != nullptr) {
			if (root->left->val >= root->val) {
				return false;
			}
			return isValidBST(root->left);
		}
		if (root->right != nullptr) {
			if (root->right->val <= root->val) {
				return false;
			}
			return isValidBST(root->right)            ;
		}
		return true;
	}
};

无法处理子树元素大于上一层的问题。中序遍历排序：

/**
 * 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) {
		vector<int> ret;
		stack<TreeNode*> st;
		TreeNode* cur;
		cur = root;
		while (!st.empty() || cur != nullptr) {
			while (cur != nullptr) {
				st.push(cur);
				cur = cur->left;
			}
			cur = st.top();
			st.pop();
			ret.push_back(cur->val);
			cur = cur->right;
		}

		int prev = ret[0];
		for (int i = 1; i < ret.size(); i++) {
			auto r = ret[i];
			if (r <= prev) {
				return false;
			}
			prev = r;
		}
		return true;
	}
};

可以去掉中间结果存数组：

/**
 * 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*> st;
		TreeNode* cur = nullptr;
		TreeNode* prev = nullptr;
		cur = root;
		while (!st.empty() || cur != nullptr) {
			while (cur != nullptr) {
				st.push(cur);
				cur = cur->left;
			}
			cur = st.top();
			st.pop();
			if (prev != nullptr && cur->val <= prev->val) {
				return false;
			}
			prev = cur;

			cur = cur->right;
		}
		return true;
	}
};

[2,1,3]
[5,1,4,null,null,3,6]
[2,2,2]
[5,4,6,null,null,3,7]