257. Binary Tree Paths

 

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"]

 

这道题给我们一个二叉树,让返回所有根到叶节点的路径,跟之前那道 Path Sum II 很类似,比那道稍微简单一些,不需要计算路径和,只需要无脑返回所有的路径即可,那么思路还是用递归来解,博主之前就强调过,玩树的题目,十有八九都是递归,而递归的核心就是不停的 DFS 到叶结点,然后在回溯回去。在递归函数中,当遇到叶结点的时候,即没有左右子结点,那么此时一条完整的路径已经形成了,加上当前的叶结点后存入结果 res 中,然后回溯。注意这里结果 res 需要 reference,而 out 是不需要引用的,不然回溯回去还要删除新添加的结点,很麻烦。为了减少判断空结点的步骤,我们在调用递归函数之前都检验一下非空即可,代码而很简洁,参见如下:

 

解法一:

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root) helper(root, "", res);
        return res;
    }
    void helper(TreeNode* node, string out, vector<string>& res) {
        if (!node->left && !node->right) res.push_back(out + to_string(node->val));
        if (node->left) helper(node->left, out + to_string(node->val) + "->", res);
        if (node->right) helper(node->right, out + to_string(node->val) + "->", res);
    }
};

 

下面再来看一种递归的方法,这个方法直接在一个函数中完成递归调用,不需要另写一个 helper 函数,核心思想和上面没有区别,参见代码如下:

 

解法二:

class Solution {
public:
    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;
    }
};

 

还是递归写法,从论坛中扒下来的解法,核心思路都一样啦,写法各有不同而已,参见代码如下:

 

解法三:

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        if (!root) return {};
        if (!root->left && !root->right) return {to_string(root->val)};
        vector<string> res;
        for (string str : binaryTreePaths(root->left)) {
            res.push_back(to_string(root->val) + "->" + str);
        }
        for (string str : binaryTreePaths(root->right)) {
            res.push_back(to_string(root->val) + "->" + str);
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/257

 

类似题目:

Path Sum II

 

参考资料:

https://leetcode.com/problems/binary-tree-paths/

https://leetcode.com/problems/binary-tree-paths/discuss/68258/Accepted-Java-simple-solution-in-8-lines

https://leetcode.com/problems/binary-tree-paths/discuss/68282/Clean-Java-solution-(Accepted)-without-any-helper-recursive-function

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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