<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 {
public:
    vector<pair<int,TreeNode*>> ancestors;
    vector<int> ans;
    bool findancestors(TreeNode* root, TreeNode* target){
        if(!root)return false;
        if(root==target){ancestors.push_back({-1,root});return true;}
        bool right = findancestors(root->right,target);
        bool left = findancestors(root->left,target);
        if(left)ancestors.push_back({1,root}); // 1 to signify left subtree
        if(right)ancestors.push_back({0,root});//0 to signify right subtree
        return right || left;
        
    }

    void findans(TreeNode* root, int k,int depth,int child){
        if(!root)return;
        if(k == depth){
            ans.push_back(root->val);
            return;
        }
        //if target found in left subtree, search paths in right subtree and vice-versa
        if(depth == 0 && child == 1){
            findans(root->right,k,depth+1,child);
        }
        else if(depth == 0 && child == 0){
            findans(root->left,k,depth+1,child);
        }
        else{
            findans(root->left,k,depth+1,child);
            findans(root->right,k,depth+1,child);
        }
    }

    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        findancestors(root,target);
        for(auto [child,ptr]: ancestors){
            findans(ptr,k,0,child);
            k--;
        }
        return ans;
    }
};
13500