[leetcode]Validate Binary Search Tree (判断有效二叉搜索树 C语言实现)

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.
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.
题意：判断一棵树是否满足二叉搜索树。
二叉搜索树的特点：左子树<根节点<右子树
解题思路：中序遍历，链栈，来实现。
对于一颗二叉树，中序遍历的结果若满足递增排序就满足二叉搜索树的条件，例如：
一颗二叉树如下：

中序遍历的结果如下：1234567

正好满足二叉搜索树的条件，所以本题采用中序遍历的思路，把遍历的结果存储于链栈中，通过比较前后存入链栈的大小，来判断是否为二叉搜索树。
实现C代码如下：

/** *解题思路：中序遍历，链栈 * Definition for binary tree * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right;  * }; */int flag;struct Node{//创建一个链栈节点    int val;    struct Node *next;};struct Stack{//创建一个链栈    struct Node *top;//指向链栈栈顶节点    int count;//记录链栈的节点个数};void InitStack(struct Stack *stack){//初始化一个空栈    stack->count = 0;    stack->top = NULL;}void PushStack(struct Stack *stack,int val){//压栈    struct Node *node;    node = (struct Node *)malloc(sizeof(struct Node));    if(stack->count > 0){        if(stack->top->val < val){//若不是第一个进栈的节点，则判断与栈顶节点的值大小，若小于栈顶节点值则说明不是二叉搜索树            node->val = val;            node->next = stack->top;            stack->top = node;            stack->count++;        }else{            flag = -1;//若不是二叉搜索树设置全局标志位flag为-1;            return;        }    }else{//第一个值进栈        node->val = val;        node->next = stack->top;        stack->top = node;        stack->count++;    }}void Inorder(struct TreeNode *root,struct Stack *stack){//中序遍历    if(root == NULL){        return;    }    Inorder(root->left,stack);    PushStack(stack,root->val);    Inorder(root->right,stack);}bool isValidBST(struct TreeNode *root) {    flag = 0;    struct Stack *stack;    stack = (struct Stack *)malloc(sizeof(struct Stack));    InitStack(stack);    Inorder(root,stack);    if(flag == -1){        return 0;    }    return 1;}