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.

思路

1. 用一個(gè)ResultStruct, 包含：
   • isBST：表示以當(dāng)前node為根節(jié)點(diǎn)的樹是否為BST：
   • minValue：表示以當(dāng)前node為根節(jié)點(diǎn)的樹，其最小的node值
   • maxValue：表示以當(dāng)前node為根節(jié)點(diǎn)的樹，其最大的node值
2. 從root開始遍歷，每個(gè)節(jié)點(diǎn)均返回這個(gè)結(jié)構(gòu)：
   • root 為空時(shí)，其結(jié)構(gòu)為（true, Integer. MAX_VALUE, Integer.MIN_VALUE）
   • 向下遍歷其左右子節(jié)點(diǎn)，看其左右子節(jié)點(diǎn)返回的結(jié)構(gòu)，是否有isBST == false，如果有，返回結(jié)構(gòu)為（false，0，0）
   • 如果其左child的maxValue >= root.val， 或者右child的minValue <= root.val，說(shuō)明當(dāng)前節(jié)點(diǎn)不是BST，同樣返回結(jié)構(gòu)為（false，0，0）
   • 如果以上都不滿足，那么說(shuō)明其是一個(gè)BST，返回結(jié)構(gòu)為true，同時(shí)返回其最大和最小node的值

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private class helpStruct {
        private boolean isBST;
        private int min;
        private int max;
        public helpStruct(boolean isBST, int min, int max) {
            this.isBST = isBST;
            this.min = min;
            this.max = max;
        }
    }
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root).isBST;
    }
    
    private helpStruct helper(TreeNode root) {
        if (root == null) {
            return new helpStruct(true, Integer.MAX_VALUE, Integer.MIN_VALUE);
        }
        
        helpStruct left = helper(root.left);
        helpStruct right = helper(root.right);
        
        if (!left.isBST || !right.isBST) {
            return new helpStruct(false, 0, 0);
        }
        
        if ((root.left != null && left.max >= root.val) || (root.right != null && right.min <= root.val)) {
            return new helpStruct(false, 0, 0);
        }
        
        return new helpStruct(true, Math.min(root.val, left.min), Math.max(root.val, right.max)); 
    }
}