Binary Tree Paths leetcode

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.

 

很简单的题目，二叉树的遍历，为了追求挑战性，我选择非递归实现

vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> out;
    if (root == nullptr)
        return out;
    vector<TreeNode*> sta;
    sta.push_back(root);
    TreeNode* lastRoot = root;
    while (!sta.empty())
    {
        root = sta.back();
        if (lastRoot != root->right)
        {
            if (lastRoot != root->left) {
                if (root->left != nullptr) {
                    sta.push_back(root->left);
                    continue;
                }
            }
            if (root->right != nullptr) {
                sta.push_back(root->right);
                continue;
            }
            else if (root->left == nullptr)
            {
                string str(to_string(sta[0]->val));
                for (int i = 1; i < sta.size(); ++i)
                    str.append("->" + to_string(sta[i]->val));
                out.push_back(str);
            }
        }
        lastRoot = root;
        sta.pop_back();
    }
    return out;
}

至于DFS深度遍历，还是老老实实的使用visited标记记录每个节点吧。上面的方法只适用于二叉树