Question

2385. Amount of Time for Binary Tree to Be Infected

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Description

You are given the root of a binary tree with unique values, and an integer start. At minute 0, an infection starts from the node with value start.

Each minute, a node becomes infected if:

• The node is currently uninfected.
• The node is adjacent to an infected node.

Return _the number of minutes needed for the entire tree to be infected._

 

Example 1:

Input: root = [1,5,3,null,4,10,6,9,2], start = 3
Output: 4
Explanation: The following nodes are infected during:
- Minute 0: Node 3
- Minute 1: Nodes 1, 10 and 6
- Minute 2: Node 5
- Minute 3: Node 4
- Minute 4: Nodes 9 and 2
It takes 4 minutes for the whole tree to be infected so we return 4.

Example 2:

Input: root = [1], start = 1
Output: 0
Explanation: At minute 0, the only node in the tree is infected so we return 0.

 

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 1 <= Node.val <= 105
• Each node has a unique value.
• A node with a value of start exists in the tree.

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
    def dfs(root):
      if root is None:
        return
      if root.left:
        g[root.val].append(root.left.val)
        g[root.left.val].append(root.val)
      if root.right:
        g[root.val].append(root.right.val)
        g[root.right.val].append(root.val)
      dfs(root.left)
      dfs(root.right)

    g = defaultdict(list)
    dfs(root)
    vis = set()
    q = deque([start])
    ans = -1
    while q:
      ans += 1
      for _ in range(len(q)):
        i = q.popleft()
        vis.add(i)
        for j in g[i]:
          if j not in vis:
            q.append(j)
    return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private Map<Integer, List<Integer>> g = new HashMap<>();

  public int amountOfTime(TreeNode root, int start) {
    dfs(root);
    Deque<Integer> q = new ArrayDeque<>();
    Set<Integer> vis = new HashSet<>();
    q.offer(start);
    int ans = -1;
    while (!q.isEmpty()) {
      ++ans;
      for (int n = q.size(); n > 0; --n) {
        int i = q.pollFirst();
        vis.add(i);
        if (g.containsKey(i)) {
          for (int j : g.get(i)) {
            if (!vis.contains(j)) {
              q.offer(j);
            }
          }
        }
      }
    }
    return ans;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    if (root.left != null) {
      g.computeIfAbsent(root.val, k -> new ArrayList<>()).add(root.left.val);
      g.computeIfAbsent(root.left.val, k -> new ArrayList<>()).add(root.val);
    }
    if (root.right != null) {
      g.computeIfAbsent(root.val, k -> new ArrayList<>()).add(root.right.val);
      g.computeIfAbsent(root.right.val, k -> new ArrayList<>()).add(root.val);
    }
    dfs(root.left);
    dfs(root.right);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  unordered_map<int, vector<int>> g;

  int amountOfTime(TreeNode* root, int start) {
    dfs(root);
    queue<int> q{{start}};
    unordered_set<int> vis;
    int ans = -1;
    while (q.size()) {
      ++ans;
      for (int n = q.size(); n; --n) {
        int i = q.front();
        q.pop();
        vis.insert(i);
        for (int j : g[i]) {
          if (!vis.count(j)) {
            q.push(j);
          }
        }
      }
    }
    return ans;
  }

  void dfs(TreeNode* root) {
    if (!root) return;
    if (root->left) {
      g[root->val].push_back(root->left->val);
      g[root->left->val].push_back(root->val);
    }
    if (root->right) {
      g[root->val].push_back(root->right->val);
      g[root->right->val].push_back(root->val);
    }
    dfs(root->left);
    dfs(root->right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func amountOfTime(root *TreeNode, start int) int {
  g := map[int][]int{}
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    if root.Left != nil {
      g[root.Val] = append(g[root.Val], root.Left.Val)
      g[root.Left.Val] = append(g[root.Left.Val], root.Val)
    }
    if root.Right != nil {
      g[root.Val] = append(g[root.Val], root.Right.Val)
      g[root.Right.Val] = append(g[root.Right.Val], root.Val)
    }
    dfs(root.Left)
    dfs(root.Right)
  }

  dfs(root)
  q := []int{start}
  ans := -1
  vis := map[int]bool{}
  for len(q) > 0 {
    ans++
    for n := len(q); n > 0; n-- {
      i := q[0]
      q = q[1:]
      vis[i] = true
      for _, j := range g[i] {
        if !vis[j] {
          q = append(q, j)
        }
      }
    }
  }
  return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function amountOfTime(root: TreeNode | null, start: number): number {
  const map = new Map<number, number[]>();
  const create = ({ val, left, right }: TreeNode) => {
    if (left != null) {
      map.set(val, [...(map.get(val) ?? []), left.val]);
      map.set(left.val, [...(map.get(left.val) ?? []), val]);
      create(left);
    }
    if (right != null) {
      map.set(val, [...(map.get(val) ?? []), right.val]);
      map.set(right.val, [...(map.get(right.val) ?? []), val]);
      create(right);
    }
  };
  create(root);
  const dfs = (st: number, fa: number) => {
    let res = 0;
    for (const v of map.get(st) ?? []) {
      if (v !== fa) {
        res = Math.max(res, dfs(v, st) + 1);
      }
    }
    return res;
  };
  return dfs(start, -1);
}

Solution 2

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
    def dfs(root):
      if root is None:
        return
      if root.left:
        g[root.val].append(root.left.val)
        g[root.left.val].append(root.val)
      if root.right:
        g[root.val].append(root.right.val)
        g[root.right.val].append(root.val)
      dfs(root.left)
      dfs(root.right)

    def dfs2(i, fa):
      ans = 0
      for j in g[i]:
        if j != fa:
          ans = max(ans, 1 + dfs2(j, i))
      return ans

    g = defaultdict(list)
    dfs(root)
    return dfs2(start, -1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private Map<Integer, List<Integer>> g = new HashMap<>();

  public int amountOfTime(TreeNode root, int start) {
    dfs(root);
    return dfs(start, -1);
  }

  private int dfs(int i, int fa) {
    int ans = 0;
    for (int j : g.getOrDefault(i, Collections.emptyList())) {
      if (j != fa) {
        ans = Math.max(ans, 1 + dfs(j, i));
      }
    }
    return ans;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    if (root.left != null) {
      g.computeIfAbsent(root.left.val, k -> new ArrayList<>()).add(root.val);
      g.computeIfAbsent(root.val, k -> new ArrayList<>()).add(root.left.val);
    }
    if (root.right != null) {
      g.computeIfAbsent(root.right.val, k -> new ArrayList<>()).add(root.val);
      g.computeIfAbsent(root.val, k -> new ArrayList<>()).add(root.right.val);
    }
    dfs(root.left);
    dfs(root.right);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  unordered_map<int, vector<int>> g;

  int amountOfTime(TreeNode* root, int start) {
    dfs(root);
    return dfs(start, -1);
  }

  int dfs(int i, int fa) {
    int ans = 0;
    for (int& j : g[i]) {
      if (j != fa) {
        ans = max(ans, 1 + dfs(j, i));
      }
    }
    return ans;
  }

  void dfs(TreeNode* root) {
    if (!root) return;
    if (root->left) {
      g[root->val].push_back(root->left->val);
      g[root->left->val].push_back(root->val);
    }
    if (root->right) {
      g[root->val].push_back(root->right->val);
      g[root->right->val].push_back(root->val);
    }
    dfs(root->left);
    dfs(root->right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func amountOfTime(root *TreeNode, start int) int {
  g := map[int][]int{}
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    if root.Left != nil {
      g[root.Val] = append(g[root.Val], root.Left.Val)
      g[root.Left.Val] = append(g[root.Left.Val], root.Val)
    }
    if root.Right != nil {
      g[root.Val] = append(g[root.Val], root.Right.Val)
      g[root.Right.Val] = append(g[root.Right.Val], root.Val)
    }
    dfs(root.Left)
    dfs(root.Right)
  }

  var dfs2 func(int, int) int
  dfs2 = func(i, fa int) int {
    ans := 0
    for _, j := range g[i] {
      if j != fa {
        ans = max(ans, 1+dfs2(j, i))
      }
    }
    return ans
  }

  dfs(root)
  return dfs2(start, -1)
}