<p>Given the <code>root</code> of a binary tree, the value of a target node <code>target</code>, and an integer <code>k</code>, return <em>an array of the values of all nodes that have a distance </em><code>k</code><em> from the target node.</em></p>

<p>You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png" style="width: 500px; height: 429px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
<strong>Output:</strong> [7,4,1]
Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1], target = 1, k = 3
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 500]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 500</code></li>
	<li>All the values <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>target</code> is the value of one of the nodes in the tree.</li>
	<li><code>0 &lt;= k &lt;= 1000</code></li>
</ul>

/**
 * 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 {
    vector<int> result; 
    vector<int> s;
    bool q = true;
public:
    int solve(TreeNode* root, int depth, const int target) {
        if (root == nullptr) {
            return 0;
        }
        if (root->val != target && q) {
            if (s.size() < depth + 1) {
                s.push_back(root->val);
            } else {
                s[depth] = root->val;
            }
            solve(root->left, depth + 1, target);
            solve(root->right, depth + 1, target);     
        } else if (root->val == target && q) {
            if (s.size() < depth + 1) {
                s.push_back(root->val);
            } else {
                s[depth] = root->val;
                s.erase(s.begin() + depth + 1, s.end());
            }
            s.erase(s.begin());
            q = false;
        }

        return 0;
    }

    int solveTwo(TreeNode* root, int depth, int distance, const int k, const int rootVal) {
        if (root == nullptr) {
            return 0;
        }
        
        if (root->val != rootVal) {
            if (depth < s.size() && s[depth] == root->val) {
                distance -= 1;
            } else {
                distance += 1;
            }
        }
        
        if (distance == k)
            result.push_back(root->val);

        solveTwo(root->left, depth + 1, distance, k, rootVal);
        solveTwo(root->right, depth + 1, distance, k, rootVal);

        return 0;
    }
    
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        solve(root, 0, target->val);
        solveTwo(root, -1, s.size(), k, root->val);
        
        return result;
    }
};
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