[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;
}