98. Validate Binary Search Tree

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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。

class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root==null) return true;
        TreeNode cur = root;
        TreeNode pre = null;
        Stack<TreeNode> stack  = new Stack<TreeNode>();
        while(cur!=null || !stack.isEmpty()){
            while(cur!=null){
            stack.push(cur);
            cur = cur.left;
            }
            cur = stack.pop();
            if(pre!=null && pre.val>=cur.val) return false;
            pre =cur;
            cur=cur.right;
        }
        return true;
        
    }
  
}

 

 

利用递归，需要将最大值最小值传下去。

class Solution {
    public boolean isValidBST(TreeNode root) {
        return test(root,Long.MIN_VALUE,Long.MAX_VALUE); 
    }
    private boolean test(TreeNode root,long min,long max){
        if(root==null) return true;
        if(root.val>=max ||root.val<=min) return false;
        return test(root.left,min,root.val) && test(root.right,root.val,max);
    }
}