Formatted question description: https://leetcode.ca/all/549.html

549. Binary Tree Longest Consecutive Sequence II

Description

Given the root of a binary tree, return the length of the longest consecutive path in the tree.

A consecutive path is a path where the values of the consecutive nodes in the path differ by one. This path can be either increasing or decreasing.

For example, [1,2,3,4] and [4,3,2,1] are both considered valid, but the path [1,2,4,3] is not valid.

On the other hand, the path can be in the child-Parent-child order, where not necessarily be parent-child order.

Example 1:

Input: root = [1,2,3]
Output: 2
Explanation: The longest consecutive path is [1, 2] or [2, 1].

Example 2:

Input: root = [2,1,3]
Output: 3
Explanation: The longest consecutive path is [1, 2, 3] or [3, 2, 1].

Constraints:

The number of nodes in the tree is in the range [1, 3 * 10⁴].
-3 * 10⁴ <= Node.val <= 3 * 10⁴

Solutions

public class Binary_Tree_Longest_Consecutive_Sequence_II {


    class Solution {
        public int longestConsecutive(TreeNode root) {

            if (root == null) {
                return 0;
            }

            int res = helper(root, 1) + helper(root, -1) + 1; // +1 is increasing, -1 is decreasing
            return Math.max(res, Math.max(longestConsecutive(root.left), longestConsecutive(root.right)));
        }

        int helper(TreeNode node, int diff) {
            if (node == null) {
                return 0;
            }

            int left = 0, right = 0;
            if (node.left != null && node.val - node.left.val == diff) {
                left = 1 + helper(node.left, diff);
            }
            if (node.right != null && node.val - node.right.val == diff) {
                right = 1 + helper(node.right, diff);
            }

            return Math.max(left, right);
        }
    }
}

// OJ: https://leetcode.com/problems/binary-tree-longest-consecutive-sequence-ii/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    pair<int, int> dfs(TreeNode *root) {
        if (!root) return {0, 0};
        auto [linc, ldec] = dfs(root->left);
        auto [rinc, rdec] = dfs(root->right);
        if (root->left) {
            if (root->val != root->left->val + 1) linc = 0;
            if (root->val != root->left->val - 1) ldec = 0;
        }
        if (root->right) {
            if (root->val != root->right->val + 1) rinc = 0;
            if (root->val != root->right->val - 1) rdec = 0;
        }
        ans = max({ ans, ldec + rinc + 1, rdec + linc + 1 });
        return { max(linc, rinc) + 1, max(ldec, rdec) + 1 };
    }
public:
    int longestConsecutive(TreeNode* root) {
        dfs(root);
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def longestConsecutive(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """

    def dfs(root):
      if not root:
        return None, 0, 0, 0  # increasing length, decreasing length, global max length
      inc = dec = 1
      left, leftInc, leftDec, leftMax = dfs(root.left)
      right, rightInc, rightDec, rightMax = dfs(root.right)
      if root.val + 1 == left:
        inc = max(leftInc + 1, inc)
      if root.val - 1 == left:
        dec = max(leftDec + 1, dec)
      if root.val + 1 == right:
        inc = max(rightInc + 1, inc)
      if root.val - 1 == right:
        dec = max(rightDec + 1, dec)
      return root.val, inc, dec, max(inc + dec - 1, leftMax, rightMax, inc, dec)

    return dfs(root)[3]

549 - Binary Tree Longest Consecutive Sequence II

549. Binary Tree Longest Consecutive Sequence II

Description

Solutions

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