All Nodes Distance K in Binary Tree in C++

Suppose we have a binary tree, a target node, and a one value K. We have to find a list of the values of all nodes that have a distance K from the target node.

So, if the input is like root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2, then the output will be [7,4,1], this is because the nodes that are a distance 2 from the target node have values 7, 4, and 1.

To solve this, we will follow these steps −

• Define a function dfs(), this will take node, pa initialize it with NULL,

• if node is null, then −

  • return

• parent[node] := pa

• dfs(left of node, node)

• dfs(right of node, node)

• From the main method do the following −

• Define an array ans

• dfs(root)

• Define one queue q for (node, value) pair

• insert { target, 0 } into q

• Define one set called visited

• insert target into visited

• while (not q is empty), do −

  • Define one pair p := first element of q

  • delete element from q

  • level := second element of temp

  • node = first element of temp.

  • if level is same as k, then −

    • insert value of node at the end of ans

  • if left of node is not null and level + 1 <= k and left of node is not visited, then

    • insert {left of node, level + 1 }) into q

    • insert left of node into visited set

  • if right of node is not null and level + 1 <= k and right of node is not visited, then

    • insert {right of node, level + 1 }) into q

    • insert right of node into visited set

  • if parent[node] is not NULL and level + 1 <= k and parent[node] is not visited, then −

    • insert { parent[node], level + 1 } into q

    • insert parent[node] into visited

• return ans

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class TreeNode{
public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = NULL;
      right = NULL;
   }
};
void insert(TreeNode **root, int val){
   queue<TreeNode*> q;
   q.push(*root);
   while(q.size()){
      TreeNode *temp = q.front();
      q.pop();
      if(!temp->left){
         if(val != NULL)
            temp->left = new TreeNode(val);
         else
            temp->left = new TreeNode(0);
         return;
      }else{
         q.push(temp->left);
      }
      if(!temp->right){
         if(val != NULL)
            temp->right = new TreeNode(val);
         else
            temp->right = new TreeNode(0);
         return;
      }else{
         q.push(temp->right);
      }
   }
}
TreeNode *make_tree(vector<int> v){
   TreeNode *root = new TreeNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      insert(&root, v[i]);
   }
   return root;
}
class Solution {
public:
   map <TreeNode*, TreeNode*> parent;
   void dfs(TreeNode* node, TreeNode* pa = NULL){
      if (!node)
         return;
      parent[node] = pa;
      dfs(node->left, node);
      dfs(node->right, node);
   }
   vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
      vector<int> ans;
      parent.clear();
      dfs(root);
      queue<pair<TreeNode*, int> > q;
      q.push({ target, 0 });
      set<TreeNode*> visited;
      visited.insert(target);
      while (!q.empty()) {
         pair<TreeNode*, int> temp = q.front();
         q.pop();
         int level = temp.second;
         TreeNode* node = temp.first;
         if (level == k) {
            ans.push_back(node->val);
         }
         if ((node->left && node->left->val != 0) && level + 1 <= k && !visited.count(node->left)) {
            q.push({ node->left, level + 1 });
            visited.insert(node->left);
         }
         if ((node->right && node->right->val != 0) && level + 1 <= k && !visited.count(node->right)){
            q.push({ node->right, level + 1 });
            visited.insert(node->right);
         }
         if (parent[node] != NULL && level + 1 <= k && !visited.count(parent[node])) {
            q.push({ parent[node], level + 1 });
            visited.insert(parent[node]);
         }
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<int> v = {3,5,1,6,2,0,8,NULL,NULL,7,4};
   TreeNode *root = make_tree(v);
   TreeNode *target = root->left;
   print_vector(ob.distanceK(root, target, 2));
}

Input

{3,5,1,6,2,0,8,NULL,NULL,7,4}

Output

[7, 4, 1, ]