【LeetCode】257. Binary Tree Paths

时间:2023-03-09 00:15:22
【LeetCode】257. Binary Tree Paths

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"]

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

深度优先遍历,每遇到叶节点,将栈中路径记录下来,最后将所有路径转成所需格式

/**
* 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:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> path;
if(root == NULL)
return path;
vector<vector<int> > pathv;
unordered_map<TreeNode*, bool> visited;
stack<TreeNode*> stk;
stk.push(root);
visited[root] = true;
if(root->left == NULL && root->right == NULL)
save(pathv, stk);
while(!stk.empty())
{
TreeNode* top = stk.top();
if(top->left && visited[top->left] == false)
{
stk.push(top->left);
visited[top->left] = true;
if(top->left->left == NULL && top->left->right == NULL)
save(pathv, stk);
continue;
}
if(top->right && visited[top->right] == false)
{
stk.push(top->right);
visited[top->right] = true;
if(top->right->left == NULL && top->right->right == NULL)
save(pathv, stk);
continue;
}
stk.pop();
}
return convert(pathv);
}
void save(vector<vector<int> >& pathv, stack<TreeNode*> stk)
{
vector<int> cur;
while(!stk.empty())
{
TreeNode* top = stk.top();
cur.push_back(top->val);
stk.pop();
}
reverse(cur.begin(), cur.end());
pathv.push_back(cur);
}
vector<string> convert(vector<vector<int> >& pathv)
{
vector<string> path;
for(int i = ; i < pathv.size(); i ++)
{
string cur;
cur += to_string(pathv[i][]);
for(int j = ; j < pathv[i].size(); j ++)
{
cur += "->";
cur += to_string(pathv[i][j]);
}
path.push_back(cur);
}
return path;
}
};

【LeetCode】257. Binary Tree Paths