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# All Nodes Distance K in Binary Tree LeetCode Solution

## Problem – All Nodes Distance K in Binary Tree

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:

``````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:

``````Input: root = , 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`

### All Nodes Distance K in Binary Tree LeetCode Solution in C++

``````class Solution {
public:
vector<int> ans;
map<TreeNode*, TreeNode*> parent;  // son=>parent
set<TreeNode*> visit;    //record visied node

void findParent(TreeNode* node){
if(!node ) return;
if( node->left ){
parent[node->left] = node;
findParent(node->left);
}
if( node->right){
parent[node->right] = node;
findParent(node->right);
}
}

vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {
if( !root ) return {};

findParent(root);
dfs(target, K );
return ans;
}
void dfs( TreeNode* node, int K){
if( visit.find(node) != visit.end() )
return;
visit.insert(node);
if( K == 0 ){
ans.push_back(node->val);
return;
}
if( node->left ){
dfs(node->left, K-1);
}
if( node->right){
dfs(node->right, K-1);
}
TreeNode* p = parent[node];
if( p )
dfs(p, K-1);
}
};
``````

### All Nodes Distance K in Binary Tree LeetCode Solution in Python

``````def distanceK(self, root, target, K):
conn = collections.defaultdict(list)
def connect(parent, child):
# both parent and child are not empty
if parent and child:
# building an undirected graph representation, assign the
# child value for the parent as the key and vice versa
conn[parent.val].append(child.val)
conn[child.val].append(parent.val)
# in-order traversal
if child.left: connect(child, child.left)
if child.right: connect(child, child.right)
# the initial parent node of the root is None
connect(None, root)
# start the breadth-first search from the target, hence the starting level is 0
bfs = [target.val]
seen = set(bfs)
# all nodes at (k-1)th level must also be K steps away from the target node
for i in range(K):
# expand the list comprehension to strip away the complexity
new_level = []
for q_node_val in bfs:
for connected_node_val in conn[q_node_val]:
if connected_node_val not in seen:
new_level.append(connected_node_val)
bfs = new_level
# add all the values in bfs into seen
seen |= set(bfs)
return bfs``````

### All Nodes Distance K in Binary Tree LeetCode Solution in Java

``````class Solution {

Map<TreeNode, Integer> map = new HashMap<>();

public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
List<Integer> res = new LinkedList<>();
find(root, target);
dfs(root, target, K, map.get(root), res);
return res;
}

// find target node first and store the distance in that path that we could use it later directly
private int find(TreeNode root, TreeNode target) {
if (root == null) return -1;
if (root == target) {
map.put(root, 0);
return 0;
}
int left = find(root.left, target);
if (left >= 0) {
map.put(root, left + 1);
return left + 1;
}
int right = find(root.right, target);
if (right >= 0) {
map.put(root, right + 1);
return right + 1;
}
return -1;
}

private void dfs(TreeNode root, TreeNode target, int K, int length, List<Integer> res) {
if (root == null) return;
if (map.containsKey(root)) length = map.get(root);
if (length == K) res.add(root.val);
dfs(root.left, target, K, length + 1, res);
dfs(root.right, target, K, length + 1, res);
}
}
``````
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