<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:
    map<TreeNode*,TreeNode*>mpp;
    void inOrder(TreeNode* root){
        if(root==NULL) return;
        if(root->left!=NULL){
            mpp[root->left]=root;
        }
        inOrder(root->left);
        if(root->right!=NULL){
            mpp[root->right]=root;
        }
        inOrder(root->right);
    }
    void BFS(TreeNode* target, int k,vector<int>&ans){
        queue<TreeNode*>q;
        q.push(target);
        set<int>visited;
        visited.insert(target->val);
        while(!q.empty()){
            int n=q.size();
            if(k==0) break;
            for(int i=0;i<n;i++){
                TreeNode* curr=q.front();
                q.pop();
                if(curr->left!=NULL && !visited.count(curr->left->val)){
                    q.push(curr->left);
                    visited.insert(curr->left->val);
                }
                if(curr->right!=NULL && !visited.count(curr->right->val)){
                    q.push(curr->right);
                    visited.insert(curr->right->val);
                }
                if(mpp.count(curr) && !visited.count(mpp[curr]->val)){
                    q.push(mpp[curr]);
                    visited.insert(mpp[curr]->val);
                }
            }
            k--;
        }
        while(!q.empty()){
            TreeNode* tmp=q.front();
            q.pop();
            ans.push_back(tmp->val);
        }
    }
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        vector<int>ans;
        inOrder(root);
        BFS(target,k,ans);
        return ans;
    }
};
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