目录
二叉查找（排序）树定义
二叉查找树插入节点
二叉查找树查找数值
二叉查找树编码与解码
LeetCode 449.Serialize and Deserialize BST
给定一个二叉查找树，实现对该二叉查找树编码与解码功能。编码即将该二叉查找树转为字符串，解码即将字符串转为二叉查找树。不限制使用何种编码算法，只需保证当对二叉查找树调用编码功能后可再调用解码功能将其复原。
总体思路
编码
解码
细节设计
编码部分
解码部分
代码实现
#include <stdio.h>
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
#include <string>
#include <vector>
void BST_insert(TreeNode *node, TreeNode *insert_node){
if (insert_node->val < node->val){
if (node->left){
BST_insert(node->left, insert_node);
}
else{
node->left = insert_node;
}
}
else{
if (node->right){
BST_insert(node->right, insert_node);
}
else{
node->right = insert_node;
}
}
}
void change_int_to_string(int val, std::string &str_val){
std::string tmp;
while(val){
tmp += val % 10 + '0';
val = val / 10;
}
for (int i = tmp.length() - 1; i >= 0; i--){
str_val += tmp[i];
}
str_val += '#';
}
void BST_preorder(TreeNode *node, std::string &data){
if (!node){
return;
}
std::string str_val;
change_int_to_string(node->val, str_val);
data += str_val;
BST_preorder(node->left, data);
BST_preorder(node->right, data);
}
class Codec {
public:
std::string serialize(TreeNode* root) {
std::string data;
BST_preorder(root, data);
return data;
}
TreeNode *deserialize(std::string data) {
if (data.length() == 0){
return NULL;
}
std::vector<TreeNode *> node_vec;
int val = 0;
for (int i = 0; i < data.length(); i++){
if (data[i] == '#'){
node_vec.push_back(new TreeNode(val));
val = 0;
}
else{
val = val * 10 + data[i] - '0';
}
}
for (int i = 1; i < node_vec.size(); i++){
BST_insert(node_vec[0], node_vec[i]);
}
return node_vec[0];
}
};
void preorder_print(TreeNode *node,int layer){
if (!node){
return;
}
for (int i = 0; i < layer; i++){
printf("-----");
}
printf("[%d]\n", node->val);
preorder_print(node->left, layer + 1);
preorder_print(node->right, layer + 1);
}
int main(){
TreeNode a(8);
TreeNode b(3);
TreeNode c(10);
TreeNode d(1);
TreeNode e(6);
TreeNode f(15);
a.left = &b;
a.right = &c;
b.left = &d;
b.right = &e;
c.left = &f;
Codec solve;
std::string data = solve.serialize(&a);
printf("%s\n", data.c_str());
TreeNode *root = solve.deserialize(data);
preorder_print(root, 0);
return 0;
}逆序数
LeetCode 315.Count of Smaller Numbers After Self
已知数组nums,求新数组count,count[i]代表了在nums[i]右侧且比nums[i]小的元素个数。
总体思路
细节设计
代码实现
#include <stdio.h>
#include <vector>
struct BSTNode {
int val;
int count;
BSTNode *left;
BSTNode *right;
BSTNode(int x) : val(x), left(NULL), right(NULL), count(0) {}
};
void BST_insert(BSTNode *node, BSTNode *insert_node, int &count_small){
if (insert_node->val <= node->val){
node->count++;
if (node->left){
BST_insert(node->left, insert_node, count_small);
}
else{
node->left = insert_node;
}
}
else{
count_small += node->count + 1;
if (node->right){
BST_insert(node->right, insert_node, count_small);
}
else{
node->right = insert_node;
}
}
}
class Solution {
public:
std::vector<int> countSmaller(std::vector<int>& nums) {
std::vector<int> result;
std::vector<BSTNode *> node_vec;
std::vector<int> count;
for (int i = nums.size() - 1; i >= 0; i--){
node_vec.push_back(new BSTNode(nums[i]));
}
count.push_back(0);
for (int i = 1; i < node_vec.size(); i++){
int count_small = 0;
BST_insert(node_vec[0], node_vec[i], count_small);
count.push_back(count_small);
}
for (int i = node_vec.size() - 1; i >= 0; i--){
delete node_vec[i];
result.push_back(count[i]);
}
return result;
}
};
int main(){
int test[] = {5, -7, 9, 1, 3, 5, -2, 1};
std::vector<int> nums;
for (int i = 0; i < 8; i++){
nums.push_back(test[i]);
}
Solution solve;
std::vector<int> result = solve.countSmaller(nums);
for (int i = 0; i < result.size(); i++){
printf("[%d]", result[i]);
}
printf("\n");
return 0;
}