LeetCode //C - 257. Binary Tree Paths

257. Binary Tree Paths

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.
 

Example 1:

Input: root = [1,2,3,null,5]
Output: [“1->2->5”,“1->3”]

Example 2:

Input: root = [1]
Output: [“1”]

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• -100 <= Node.val <= 100

From: LeetCode
Link: 257. Binary Tree Paths

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Solution:

Ideas:

1. TreeNode Structure: The structure for a binary tree node is defined with integer value, left child, and right child pointers.

2. DFS Helper Function: The dfs function performs a depth-first search to find all paths from the root to the leaves. It constructs the path as a string and adds it to the paths array when a leaf is encountered.

3. Main Function: The binaryTreePaths function initializes the paths array and calls the dfs function to populate it. It also sets the return size of the paths array.

4. Helper Functions:

• newNode creates a new tree node.
• freeTree frees the allocated tree nodes.
• freePaths frees the allocated paths array.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
void dfs(struct TreeNode* root, char* path, int length, char** paths, int* size) {
    if (root == NULL) return;
    
    int len = snprintf(path + length, 100 - length, "%d", root->val);
    length += len;
    
    if (root->left == NULL && root->right == NULL) {
        paths[*size] = (char*)malloc((length + 1) * sizeof(char));
        strcpy(paths[*size], path);
        (*size)++;
    } else {
        length += snprintf(path + length, 100 - length, "->");
        dfs(root->left, path, length, paths, size);
        dfs(root->right, path, length, paths, size);
    }
}

char** binaryTreePaths(struct TreeNode* root, int* returnSize) {
    char** paths = (char**)malloc(100 * sizeof(char*));
    char path[100];
    *returnSize = 0;
    
    if (root != NULL) {
        dfs(root, path, 0, paths, returnSize);
    }
    
    return paths;
}

// Helper function to create a new tree node
struct TreeNode* newNode(int data) {
    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node->val = data;
    node->left = NULL;
    node->right = NULL;
    return node;
}

// Helper function to free the allocated tree nodes
void freeTree(struct TreeNode* root) {
    if (root == NULL) return;
    freeTree(root->left);
    freeTree(root->right);
    free(root);
}

// Helper function to free the paths array
void freePaths(char** paths, int size) {
    for (int i = 0; i < size; i++) {
        free(paths[i]);
    }
    free(paths);
}

原文地址：https://blog.csdn.net/navicheung/article/details/140624855

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