Leetcode

Maximum Width of Binary Tree

May 26, 2022 by aAshish

Approach 1: BFS

  • Time:O(n)
  • Space:O(h)

C++

 
class Solution {
 public:
  int widthOfBinaryTree(TreeNode* root) {
    if (!root)
      return 0;

    int ans = 0;
    queue<pair<TreeNode*, int>> q{{{root, 1}}};  // {node, index}

    while (!q.empty()) {
      const int offset = q.front().second * 2;
      ans = max(ans, q.back().second - q.front().second + 1);
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [node, index] = q.front();
        q.pop();
        if (node->left)
          q.emplace(node->left, index * 2 - offset);
        if (node->right)
          q.emplace(node->right, index * 2 + 1 - offset);
      }
    }

    return ans;
  }
};

JAVA

 
class Solution {
  public int widthOfBinaryTree(TreeNode root) {
    if (root == null)
      return 0;

    int ans = 0;
    Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>(); // {node, index}
    q.offer(new Pair<>(root, 1));

    while (!q.isEmpty()) {
      final int offset = q.peekFirst().getValue() * 2;
      ans = Math.max(ans, q.peekLast().getValue() - q.peekFirst().getValue() + 1);
      for (int sz = q.size(); sz > 0; --sz) {
        final TreeNode node = q.peekFirst().getKey();
        final int index = q.pollFirst().getValue();
        if (node.left != null)
          q.offer(new Pair<>(node.left, index * 2 - offset));
        if (node.right != null)
          q.offer(new Pair<>(node.right, index * 2 + 1 - offset));
      }
    }

    return ans;
  }
}

Approach 2: DFS (Overflow)

  • Time:O(n)
  • Space:O(n)

C++

 
class Solution {
 public:
  int widthOfBinaryTree(TreeNode* root) {
    if (!root)
      return 0;

    long ans = 0;
    dfs(root, 0, 1, {}, ans);
    return ans;
  }

 private:
  void dfs(TreeNode* root, int level, long index, vector<long>&& startOfLevel,
           long& ans) {
    if (!root)
      return;
    if (startOfLevel.size() == level)
      startOfLevel.push_back(index);

    ans = max(ans, index - startOfLevel[level] + 1);
    dfs(root->left, level + 1, index * 2, move(startOfLevel), ans);
    dfs(root->right, level + 1, index * 2 + 1, move(startOfLevel), ans);
  }
};