98. Validate Binary Search Tree
2016-09-27 23:57
316 查看
验证二叉搜索树的正确性:
法一:二叉搜索树的中序遍历是递增数列
法二:
递归判断每一个子树
原理:可以证明:一个树是BST <=> 这个树的左孩子是BST&&这个树的右孩子是BST&&root大于所有左孩子&&root小于所有右孩子
法一:二叉搜索树的中序遍历是递增数列
public class Solution { public List<Integer> inorderTraversal(TreeNode root){ List<Integer> result = new ArrayList<Integer>(); Stack<TreeNode> nodeStack = new Stack<TreeNode>(); while(root != null || !nodeStack.empty()){ if (root != null){ nodeStack.push(root); root = root.left; }else{ root = nodeStack.pop(); result.add(root.val); root = root.right; } } return result; } public boolean isValidBST(TreeNode root) { List<Integer> list = inorderTraversal(root); for (int i = 0; i < list.size() - 1; i ++){ if (list.get(i) >= list.get(i+1)) return false; } return true; } }
法二:
递归判断每一个子树
原理:可以证明:一个树是BST <=> 这个树的左孩子是BST&&这个树的右孩子是BST&&root大于所有左孩子&&root小于所有右孩子
public boolean isValidBST(TreeNode root) { if (root.val > root.left.max_value() && root < root.right.min_value()) return isValidBST(root.left) && isValidBST(root.right); else return false; } 其中,root.left.max_value()就等于root.left的最左边的孩子;root.right.min_value()等于最右边的孩子;
相关文章推荐
- [Leetcode] 98. Validate Binary Search Tree
- leetcode 98. Validate Binary Search Tree
- *LeetCode 98. Validate Binary Search Tree
- 第十八周:[Leetcode]98. Validate Binary Search Tree
- [LeetCode] 98. Validate Binary Search Tree
- Leetcode 98. Validate Binary Search Tree
- leetcode 98. Validate Binary Search Tree
- 98. Validate Binary Search Tree
- leetcode 98. Validate Binary Search Tree DFS深度优先搜索 + 两个递归函数 + 一个错误做法
- 98. Validate Binary Search Tree
- [LeetCode]98. Validate Binary Search Tree
- 98. Validate Binary Search Tree
- LeetCode 98. Validate Binary Search Tree
- [Leetcode] 98. Validate Binary Search Tree @python
- LeetCode 98. Validate Binary Search Tree
- leetcode oj java 98. Validate Binary Search Tree
- LeetCode 98. Validate Binary Search Tree
- 个人记录-LeetCode 98. Validate Binary Search Tree
- 98. Validate Binary Search Tree
- Leetcode 98. Validate Binary Search Tree