【Leecode】207Validate Binary Search Tree有效二叉搜索树

/*
2.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.
*/

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
判断一个二叉搜索树是否有效，左边子树的结点小于结点值，右边子树的结点大于结点值
思路：中序遍历
我们利用二分查找的性质，二分查找树的中序遍历是按照顺序递增的。所以我们只需要中序遍历一遍这棵树，维护前驱结点，每次检查是否满足递增关系即可。我们先中序遍历左孩子，直到没有左孩子，返回和前驱结点比较，然后递归这个节点的右孩子。即完成中序遍历过程。
Attention:
1. 前驱结点需要设置为引用类型，因为我们需要这个pre的值能够影响它在函数外部的值，即在各层递归中都同时改变。

bool in_order(TreeNode* node, TreeNode* &pre) 

http://blog.csdn.net/cinderella_niu/article/details/43115689

class Solution {public:    bool isValidBST(TreeNode* root) {        TreeNode* pre = NULL;        return middle_order(root,pre);    }private:    bool middle_order(TreeNode* node, TreeNode* &pre)    {        if(node == NULL)            return true;        if(!middle_order(node->left,pre))             return false;        if(pre!=NULL && pre->val >= node->val)            return false;        pre = node;        return middle_order(node->right, pre);    }};

版本二：java代码，来自《程序员代码面试指南IT名企算法与数据结构题目最优解—左程云著》

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isValidBST(TreeNode root) {        if (root == null) {            return true;        }        boolean res = true;        TreeNode pre = null;        TreeNode cur1 = root;        TreeNode cur2 = null;        while (cur1 != null) {            cur2 = cur1.left;            if (cur2 != null) {                while (cur2.right != null && cur2.right != cur1) {                    cur2 = cur2.right;                }                if (cur2.right == null) {                    cur2.right = cur1;                    cur1 = cur1.left;                    continue;                } else {                    cur2.right = null;                }            }            if (pre != null && pre.val >= cur1.val) {                res = false;            }            pre = cur1;            cur1 = cur1.right;        }        return res;    }}