98. Validate Binary Search Tree

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:

    2
   / 
  1   3

Binary tree [2,1,3], return true.

Example 2:

    1
   / 
  2   3

Binary tree [1,2,3], return false.

利用中序遍历，因为中序遍历的结果就是从小到大排好序的结果，如果是二叉搜索树的话。

 1 class Solution {
 2     public boolean isValidBST(TreeNode root) {
 3         if(root==null) return true;
 4         TreeNode cur = root;
 5         TreeNode pre = null;
 6         Stack<TreeNode> stack  = new Stack<TreeNode>();
 7         while(cur!=null || !stack.isEmpty()){
 8             while(cur!=null){
 9             stack.push(cur);
10             cur = cur.left;
11             }
12             cur = stack.pop();
13             if(pre!=null && pre.val>=cur.val) return false;
14             pre =cur;
15             cur=cur.right;
16         }
17         return true;
18         
19     }
20   
21 }

利用递归，需要将最大值最小值传下去。

 1 class Solution {
 2     public boolean isValidBST(TreeNode root) {
 3         return test(root,Long.MIN_VALUE,Long.MAX_VALUE); 
 4     }
 5     private boolean test(TreeNode root,long min,long max){
 6         if(root==null) return true;
 7         if(root.val>=max ||root.val<=min) return false;
 8         return test(root.left,min,root.val) && test(root.right,root.val,max);
 9     }
10 }