Problem

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.
A single node tree is a BST

Example

An example:

  2
 / 
1   4
   / 
  3   5

The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).

Note

第一种，先跑左子树的DFS：如果不满足，返回false.
左子树分支判断完成后，pre = root.left（因为唯一给pre赋值的语句是pre = root，所以运行完dfs(root.left)之后，pre = root.left）。
然后，若pre.val，也就是root.left.val，大于等于root.val的话，不满足BST定义，也返回false；
否则，继续执行pre = root，（现在去判断root.right，所以用root.val和root.right.val比较。）不满足就返回false.
如果跑完了右子树的DFS，还没有返回false，返回true.
第二种DFS，用root的值给左右子树的递归设定边界。

Solution

DFS I

public class Solution {
    TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        return dfs(root);   
    }
    public boolean dfs(TreeNode root) {
        if (root == null) return true;
        if (!dfs(root.left)) return false;
        if (pre != null && pre.val >= root.val) return false;
        pre = root;
        if (!dfs(root.right)) return false;
        return true;
    }

}

DFS II

class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValid(root, null, null);
    }
    private boolean isValid(TreeNode node, Integer min, Integer max) {
        if (node == null) return true;
        if (min != null && node.val <= min) return false;
        if (max != null && node.val >= max) return false;
        return isValid(node.left, min, node.val) && isValid(node.right, node.val, max);
    }
}