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 = [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

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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