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298. Binary Tree Longest Consecutive Sequence 👍

  • Time: $O(n)$
  • Space: $O(h)$
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class Solution {
 public:
  int longestConsecutive(TreeNode* root) {
    if (root == nullptr)
      return 0;
    return dfs(root, root->val, 0, 0);
  }

 private:
  int dfs(TreeNode* root, int target, int length, int maxLength) {
    if (root == nullptr)
      return maxLength;
    if (root->val == target)
      maxLength = max(maxLength, ++length);
    else
      length = 1;
    return max(dfs(root->left, root->val + 1, length, maxLength),
               dfs(root->right, root->val + 1, length, maxLength));
  }
};

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class Solution {
  public int longestConsecutive(TreeNode root) {
    if (root == null)
      return 0;
    return dfs(root, -1, 0, 1);
  }

  private int dfs(TreeNode root, int target, int length, int mx) {
    if (root == null)
      return mx;
    if (root.val == target)
      mx = Math.max(mx, ++length);
    else
      length = 1;
    return Math.max(dfs(root.left, root.val + 1, length, mx),
                    dfs(root.right, root.val + 1, length, mx));
  }
}

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class Solution:
  def longestConsecutive(self, root: Optional[TreeNode]) -> int:
    if not root:
      return 0

    def dfs(root: Optional[TreeNode], target: int, length: int, maxLength: int) -> int:
      if not root:
        return maxLength
      if root.val == target:
        length += 1
        maxLength = max(maxLength, length)
      else:
        length = 1
      return max(dfs(root.left, root.val + 1, length, maxLength),
                 dfs(root.right, root.val + 1, length, maxLength))

    return dfs(root, root.val, 0, 0)