二叉排序樹排序規則： 左節點 <= 根節點 <= 右節點

樹節點：

typedef struct _node
{
    int data;
    struct _node *leftChild;
    struct _node *rightChild;
}TreeNode;

操作：

//按照儲存規則插入節點
void insertNode(TreeNode **root, int data)
{
    TreeNode *node = (TreeNode *) malloc(sizeof(TreeNode));
    node->data = data;
    node->
 
leftChild = NULL;
    node->rightChild = NULL;
    //1.樹為空
    //2.樹不為空
    if (*root == NULL)
    {
        *root = node;
    }
    else
    {
        TreeNode *current = *root;
        while (1)
        {
            if (current->data >= data)
            {
                if (current->leftChild ==
 
 NULL)
                {
                    current->leftChild = node;
                    break;
                }
                else current = current->leftChild;
            }
            else
            {
                if (current->rightChild == NULL)
                {
                    current->
 
rightChild = node;
                    break;
                }
                else current = current->rightChild;
            }
        }
    }
}


//中序遍歷：左->根->右，按照從小到大順序輸出樹
void inOrder(TreeNode *node)
{
    if(node != NULL)
    {
        inOrder(node->leftChild);
        printf("%d ",node->data);
        inOrder(node->rightChild);
    }
}

void endl()
{
    printf("\n");
}


//按照後序遍歷(PostOrder)(左->右->根)的順序釋放節點記憶體
void freeAll(TreeNode *node)
{
    if(node->leftChild != NULL) freeAll(node->leftChild);
    if(node->rightChild != NULL) freeAll(node->rightChild);
    free(node);
}

int main()
{
    TreeNode *rootNode = NULL;
    insertNode(&rootNode, 1);
    insertNode(&rootNode, 2);
    insertNode(&rootNode, 3);
    insertNode(&rootNode, 2);
    insertNode(&rootNode, 3);
    insertNode(&rootNode, 98);
    insertNode(&rootNode, 13);
    inOrder(rootNode);
    endl();
    insertNode(&rootNode, 6);
    insertNode(&rootNode, 13);
    insertNode(&rootNode, 6);
    insertNode(&rootNode, 54);
    insertNode(&rootNode, 44);

    inOrder(rootNode);
    endl();

    freeAll(rootNode);
    return 0;
}

小結：由於二叉排序樹的規則，插入的節點保證嚴格有序，節點查詢的時間複雜度在O(Log2N)~O(N)之間，故整體排序的時間複雜度在O(NLog2N)~O(N^2)之間