# All Nodes Distance K In Binary Tree Problem

## Description

LeetCode Problem 863.

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

You can return the answer in any order.

Example 1:

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Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
Output: [7,4,1]
Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1.
``````

Example 2:

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Input: root = [1], target = 1, k = 3
Output: []
``````

Constraints:

• The number of nodes in the tree is in the range [1, 500].
• 0 <= Node.val <= 500
• All the values Node.val are unique.
• target is the value of one of the nodes in the tree.
• 0 <= k <= 1000

## Sample C++ Code

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/**
* 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<TreeNode*> path(TreeNode *root, int val) {
if (!root)
return {};
if (root->val == val)
return {root};
vector<TreeNode*> l, r;
l = path(root->left, val);
r = path(root->right, val);
if (l.size()) {
l.push_back(root);
return l;
}
if (r.size()) {
r.push_back(root);
return r;
}
return {};
}

void distance(TreeNode *root, int k, TreeNode* before, vector<int> &ans) {
if (!root || root == before)
return;
distance(root->left, k - 1, before, ans);
distance(root->right, k - 1, before, ans);
if (k == 0)
ans.push_back(root->val);
return;
}

vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
vector<TreeNode*> v = path(root, target->val);
vector<int> ans;
for (int i = 0; i < v.size() && k >= i; i++)
distance(v[i], k - i, i == 0 ? NULL : v[i - 1], ans);
return ans;
}
};
``````