[LintCode] Binary Tree Paths 二叉树路径

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

Given the following binary tree:

   1
 /  
2     3
 
  5

All root-to-leaf paths are:

[
  "1->2->5",
  "1->3"
]

LeetCode上的原题,请参见我之前的博客Binary Tree Paths

解法一:

class Solution {
public:
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root) helper(root, "", res);
        return res;
    }
    void helper(TreeNode *node, string out, vector<string> &res) {
        out += to_string(node->val);
        if (!node->left && !node->right) {
            res.push_back(out);
        } else {
            if (node->left) helper(node->left, out + "->", res);
            if (node->right) helper(node->right, out + "->", res);
        }
    }
};

解法二:

class Solution {
public:
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    vector<string> binaryTreePaths(TreeNode* root) {
        if (!root) return {};
        if (!root->left && !root->right) return {to_string(root->val)};
        vector<string> left = binaryTreePaths(root->left);
        vector<string> right = binaryTreePaths(root->right);
        left.insert(left.end(), right.begin(), right.end());
        for (auto &a : left) {
            a = to_string(root->val) + "->" + a;
        }
        return left;
    }
};
原文地址:https://www.cnblogs.com/grandyang/p/5562581.html