C++ 中二叉树中所有节点的距离为 K

c++server side programmingprogramming

假设我们有一棵二叉树、一个目标节点和一个值 K。我们必须找到与目标节点距离为 K 的所有节点的值列表。

因此，如果输入为 root = [3,5,1,6,2,0,8,null,null,7,4]、target = 5、K = 2，则输出将为 [7,4,1]，这是因为距离目标节点 2 的节点的值为 7、4 和 1。

为了解决这个问题，我们将遵循以下步骤−

定义一个函数 dfs()，它将获取 node，pa 用 NULL 初始化它，
如果 node 为 null，则 −
- 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

示例

让我们看下面的实现，以便更好地理解 −

#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));
}