[LeetCode] 257. Binary Tree Paths
grandyang opened this issue · comments
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**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:
类似题目:
参考资料:
https://leetcode.com/problems/binary-tree-paths/