<LeetCode OJ> 257. Binary Tree Paths

257. Binary Tree Paths

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Total Accepted: 29282 Total
Submissions: 113527 Difficulty: Easy

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"]

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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(M) Path Sum II

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //这个算法写的太精妙了，参考讨论区，别人家的!
class Solution {
public:
    void getDfsPaths(vector<string>& result, TreeNode* node, string strpath) {
        if(!node->left && !node->right){//叶子
            result.push_back(strpath);
            return ;
        }
        if(node->left)
            getDfsPaths(result, node->left, strpath+"->"+to_string(node->left->val));
        if(node->right)
            getDfsPaths(result, node->right, strpath+"->"+to_string(node->right->val));
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ret;
        if(!root)
            return ret;

        getDfsPaths(ret, root, to_string(root->val));
        return ret;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50493936

原作者博客：http://blog.csdn.net/ebowtang