863. All Nodes Distance K in Binary Tree.md

863. All Nodes Distance K in Binary Tree

We are given a binary tree (with root node root), a target node, and an integer value K.

Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned 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.

Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Note:

• The given tree is non-empty.
• Each node in the tree has unique values 0 <= node.val <= 500.
• The target node is a node in the tree.
• 0 <= K <= 1000.

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Solution

• mine

  • Java

    Runtime: 12 ms, faster than 37.40%, Memory Usage: 39.4 MB, less than 5.26% of Java online submissions

    public Map<Integer, List<TreeNode>> map;
    public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
        LinkedList<Integer> res = new LinkedList<>();
        if (K == 0) {
            res.add(target.val);
            return res;
        }
        map = new HashMap<>();
    // key is each node, and value is a list which is connected the node 
        initMap(root, null);
    //record the node is used 
        int[] record = new int[501];
    //get the connected node
        List<TreeNode> nodes = map.get(target.val);
    //mark this node was used
        record[target.val] = 1;
        for (int i = 0; i < nodes.size(); i++) {
            record[nodes.get(i).val] = 1;
        }
    // if  K > 1 , we need to get the 
        while (K > 1) {
            List<TreeNode> temp = new LinkedList<>();
            K--;
            for (int i = 0; i < nodes.size(); i++) {
    	//get the connected nodes
                List<TreeNode> t = map.get(nodes.get(i).val);
                for (int j = 0; j < t.size(); j++) {
                    int val = t.get(j).val;
                    if (record[val] != 1) {
    		//if node is not used, save it to get next connected node.
                        record[val] = 1;
                        temp.add(t.get(j));
                    }
                }
            }
            nodes = temp;
        }
    
        for (int i = 0; i < nodes.size(); i++) {
            res.add(nodes.get(i).val);
        }
        return res;
    }
    
    public void initMap(TreeNode node, TreeNode parent) {
        if (node == null) {
            return;
        }
        List<TreeNode> t = map.getOrDefault(node.val, new ArrayList<>(3));
        if (parent != null) {
            t.add(parent);
        }
        if (node.left != null) {
            t.add(node.left);
        }
        if (node.right != null) {
            t.add(node.right);
        }
        map.put(node.val, t);
        initMap(node.left, node);
        initMap(node.right, node);
    }
    

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• the most votes

  Runtime: 10 ms, faster than 86.42%, Memory Usage: 39.6 MB, less than 5.26% of Java online submissions

  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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• the leetcode's solution

  Runtime: 9 ms, faster than 100.00%, Memory Usage: 39.1 MB, less than 5.26% of Java online submissions

  List<Integer> ans;
  TreeNode target;
  int K;
  public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
      ans = new LinkedList();
      this.target = target;
      this.K = K;
      dfs(root);
      return ans;
  }
  
  // Return vertex distance from node to target if exists, else -1
  // Vertex distance: the number of vertices on the path from node to target
  public int dfs(TreeNode node) {
      if (node == null)
          return -1;
      else if (node == target) {
          subtree_add(node, 0);
          return 1;
      } else {
          int L = dfs(node.left), R = dfs(node.right);
          if (L != -1) {
              if (L == K) ans.add(node.val);
              subtree_add(node.right, L + 1);
              return L + 1;
          } else if (R != -1) {
              if (R == K) ans.add(node.val);
              subtree_add(node.left, R + 1);
              return R + 1;
          } else {
              return -1;
          }
      }
  }
  
  // Add all nodes 'K - dist' from the node to answer.
  public void subtree_add(TreeNode node, int dist) {
      if (node == null) return;
      if (dist == K)
          ans.add(node.val);
      else {
          subtree_add(node.left, dist + 1);
          subtree_add(node.right, dist + 1);
      }
  }