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.
Example 1:
Input:
2
/
1 3
Output: true
Example 2:
5 / 1 4 / 3 6 Output: false Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value is 5 but its right child's value is 4.
验证一棵树是否为二叉搜索树
java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public boolean isValidBST(TreeNode root) { 12 if (root == null) 13 return true ; 14 Stack<TreeNode> stack = new Stack<TreeNode>() ; 15 TreeNode cur = root ; 16 TreeNode pre = null ; 17 boolean flag = true ; 18 while(cur != null || !stack.isEmpty()){ 19 while(cur != null){ 20 stack.push(cur) ; 21 cur = cur.left ; 22 } 23 TreeNode node = stack.pop() ; 24 if (pre != null && node.val <= pre.val){ 25 flag = false ; 26 break ; 27 } 28 pre = node ; 29 cur = node.right ; 30 } 31 return flag ; 32 } 33 }