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515. Find Largest Value in Each Tree Row 👍

Approach 1: BFS

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> largestValues(TreeNode* root) {
    if (root == nullptr)
      return {};

    vector<int> ans;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      int mx = INT_MIN;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode* node = q.front();
        q.pop();
        mx = max(mx, node->val);
        if (node->left)
          q.push(node->left);
        if (node->right)
          q.push(node->right);
      }
      ans.push_back(mx);
    }

    return ans;
  }
};

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class Solution {
  public List<Integer> largestValues(TreeNode root) {
    if (root == null)
      return new ArrayList<>();

    List<Integer> ans = new ArrayList<>();
    Queue<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));

    while (!q.isEmpty()) {
      int mx = Integer.MIN_VALUE;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode node = q.poll();
        mx = Math.max(mx, node.val);
        if (node.left != null)
          q.offer(node.left);
        if (node.right != null)
          q.offer(node.right);
      }
      ans.add(mx);
    }

    return ans;
  }
}

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class Solution:
  def largestValues(self, root: Optional[TreeNode]) -> List[int]:
    if not root:
      return []

    ans = []
    q = collections.deque([root])

    while q:
      mx = -math.inf
      for _ in range(len(q)):
        root = q.popleft()
        mx = max(mx, root.val)
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)
      ans.append(mx)

    return ans

Approach 2: DFS

  • Time: $O(n)$
  • Space: $O(h)$
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class Solution {
 public:
  vector<int> largestValues(TreeNode* root) {
    vector<int> ans;
    dfs(root, 0, ans);
    return ans;
  }

 private:
  void dfs(TreeNode* root, int depth, vector<int>& ans) {
    if (root == nullptr)
      return;
    if (depth + 1 > ans.size())
      ans.push_back(root->val);
    else
      ans[depth] = max(ans[depth], root->val);

    dfs(root->left, depth + 1, ans);
    dfs(root->right, depth + 1, ans);
  }
};

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class Solution {
  public List<Integer> largestValues(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    dfs(root, 0, ans);
    return ans;
  }

  private void dfs(TreeNode root, int depth, List<Integer> ans) {
    if (root == null)
      return;
    if (depth + 1 > ans.size())
      ans.add(root.val);
    else
      ans.set(depth, Math.max(ans.get(depth), root.val));

    dfs(root.left, depth + 1, ans);
    dfs(root.right, depth + 1, ans);
  }
}

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class Solution:
  def largestValues(self, root: Optional[TreeNode]) -> List[int]:
    ans = []

    def dfs(root: Optional[TreeNode], depth: int) -> None:
      if not root:
        return
      if depth + 1 > len(ans):
        ans.append(root.val)
      else:
        ans[depth] = max(ans[depth], root.val)

      dfs(root.left, depth + 1)
      dfs(root.right, depth + 1)

    dfs(root, 0)
    return ans