如何计算二叉树所有路径上的数字和
Posted on Thu 22 August 2024 in Journal
Abstract | 如何计算二叉树所有路径上的数字和 |
---|---|
Authors | Walter Fan |
Category | learning note |
Status | v1.0 |
Updated | 2024-08-22 |
License | CC-BY-NC-ND 4.0 |
拳不离口, 曲不离手
身为一个专业程序员, 一天不练习算法套路, 一天手就会生, 今天来练练下面这道题目, 忙活了好一会儿才搞定, 手确实有点生.
Question
给定一个二叉树, 求从根节点和叶子节点的所有路径所组成的数字之和
例如如下的二叉树
1
/ \
2 3
/ \
4 5
它的遍历路径为
- 124
- 125
- 13
总和为 262
Solution
废话不多说, 直接贴代码, 其实主要就是二叉树的深度遍历
#include <iostream>
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
void dfs(TreeNode* node, std::vector<int>& path, std::vector<std::string>& paths) {
if (!node) return;
// Add the current node's value to the path
path.push_back(node->val);
// If it's a leaf node, convert the path to a string and add to paths vector
if (!node->left && !node->right) {
std::string leafPath;
for (int i = 0; i < path.size(); ++i) {
//if (i > 0) leafPath += "->";
leafPath += std::to_string(path[i]);
}
paths.push_back(leafPath);
}
// Recursively traverse the left and right subtrees
dfs(node->left, path, paths);
dfs(node->right, path, paths);
// Backtrack: remove the current node's value from the path
path.pop_back();
}
std::vector<std::string> getRootToLeafPaths(TreeNode* root) {
std::vector<std::string> paths;
std::vector<int> path;
dfs(root, path, paths);
return paths;
}
TreeNode* createTree(const std::vector<int>& values, int index) {
if (index >= values.size() || values[index] == -1) {
return nullptr;
}
TreeNode* node = new TreeNode(values[index]);
node->left = createTree(values, 2 * index + 1);
node->right = createTree(values, 2 * index + 2);
return node;
}
void deleteTree(TreeNode* node) {
if (!node) return;
deleteTree(node->left);
deleteTree(node->right);
delete node;
}
int kata12_tree_path_sum(int argc, char** argv) {
// Example usage:
// Creating a binary tree with the following structure:
// 1
// / \
// 2 3
// / \
// 4 5
std::vector<int> values = {1, 2, 3, 4, 5, -1, -1};
TreeNode* root = createTree(values, 0);
// Get all root-to-leaf paths
std::vector<std::string> paths = getRootToLeafPaths(root);
int sum = 0;
for (const auto& path : paths) {
std::cout << path << std::endl;
sum += atoi(path.c_str());
}
cout << "sum=" << sum << endl;
// Clean up the tree to prevent memory leaks
deleteTree(root);
return 0;
}
本作品采用知识共享署名-非商业性使用-禁止演绎 4.0 国际许可协议进行许可。