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

本文有两种DFS方法。

先看第一种，先跑左子树的DFS：如果不满足，返回false。在左子树分支判断完成后，pre = root.left（为什么？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，（呐，现在去判断root.right了，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，耶，返回true，大功告成。

第二种DFS，用root的值给左右子树的递归设定边界，注意，要用long型的最大值和最小值哦！

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

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