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.
题意:给定一棵二叉树,推断它是不是合法的二叉查找树
思路:dfs
合法二叉查找树必须满足下面两个条件
1.左子树和右子树都是合法二叉查找树
2.左子树的最右叶子节点 < 根 < 右子树的最左叶子节点
复杂度:时间O(n),空间O(log n)
bool isValidBST(TreeNode *root) { if(!root) return true; TreeNode *right_most = root->left, *left_most = root->right; while(right_most && right_most->right){ right_most = right_most->right; } while(left_most && left_most->left){ left_most = left_most->left; } return isValidBST(root->left) && isValidBST(root->right) && (!right_most || right_most->val < root->val) && (!left_most || root->val < left_most->val); }