257. Binary Tree Paths

#week18

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   
2     3
 
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

分析：

这是一题比较简单的递归

通过递归来求解二叉树路径

题解：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void binaryTreePaths(vector<string>& result, TreeNode* root, string t) {
        if (!root->left && !root->right) {
            result.push_back(t);
            return ;
        }
        
        if (root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val));
        if (root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val));
        
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        if (!root) return result;
        
        binaryTreePaths(result, root, to_string(root->val));
        return result;
    }
};