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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);
        }
    }
}

############

/**
 * 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 int ans;

    public int longestConsecutive(TreeNode root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[] {0, 0};
        }
        int incr = 1, decr = 1;
        int[] left = dfs(root.left);
        int[] right = dfs(root.right);
        if (root.left != null) {
            if (root.left.val + 1 == root.val) {
                incr = left[0] + 1;
            }
            if (root.left.val - 1 == root.val) {
                decr = left[1] + 1;
            }
        }
        if (root.right != null) {
            if (root.right.val + 1 == root.val) {
                incr = Math.max(incr, right[0] + 1);
            }
            if (root.right.val - 1 == root.val) {
                decr = Math.max(decr, right[1] + 1);
            }
        }
        ans = Math.max(ans, incr + decr - 1);
        return new int[] {incr, decr};
    }
}

  • // 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

'''
>>> float('-inf')+1
-inf
>>> float('-inf')-1
-inf
'''
class Solution(object):
    def longestConsecutive(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        def dfs(root):
            if not root:
                # val, increasing length, decreasing length, max length
                return float('-inf'), 0, 0, 0
            # inc/dec starting from root
            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)
            # note: max may occur in children subtree, not current root's tree
            return root.val, inc, dec, max(inc + dec - 1, leftMax, rightMax, inc,
                                           dec)  # -1 because root is counted twice, even left/right both null

        return dfs(root)[3]

############

# 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 longestConsecutive(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return [0, 0]
            nonlocal ans
            incr = decr = 1
            i1, d1 = dfs(root.left)
            i2, d2 = dfs(root.right)
            if root.left:
                if root.left.val + 1 == root.val:
                    incr = i1 + 1
                if root.left.val - 1 == root.val:
                    decr = d1 + 1
            if root.right:
                if root.right.val + 1 == root.val:
                    incr = max(incr, i2 + 1) # incr1 and incr2 must be in different tree branches, eg. left child cannot be both increasing and decreasing. Same as decr1 and decr2
                if root.right.val - 1 == root.val:
                    decr = max(decr, d2 + 1)
            ans = max(ans, incr + decr - 1) # -1 because root is counted twice, even left/right both null
            return [incr, decr]

        ans = 0
        dfs(root)
        return ans


class Solution: # issue with this solution, duplicated dfs() search
    def longestConsecutive(self, root: TreeNode) -> int:
        def helper(node: TreeNode, diff: int) -> int:
            if not node:
                return 0

            left = right = 0
            if node.left and node.val - node.left.val == diff:
                left = 1 + helper(node.left, diff)
            if node.right and node.val - node.right.val == diff:
                right = 1 + helper(node.right, diff)

            return max(left, right)
        if not root:
            return 0

        res = helper(root, 1) + helper(root, -1) + 1 # +1 is increasing, -1 is decreasing
        return max(res, max(longestConsecutive(root.left), longestConsecutive(root.right)))



  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestConsecutive(root *TreeNode) int {
	ans := 0
	var dfs func(root *TreeNode) []int
	dfs = func(root *TreeNode) []int {
		if root == nil {
			return []int{0, 0}
		}
		incr, decr := 1, 1
		left := dfs(root.Left)
		right := dfs(root.Right)
		if root.Left != nil {
			if root.Left.Val+1 == root.Val {
				incr = left[0] + 1
			}
			if root.Left.Val-1 == root.Val {
				decr = left[1] + 1
			}
		}
		if root.Right != nil {
			if root.Right.Val+1 == root.Val {
				incr = max(incr, right[0]+1)
			}
			if root.Right.Val-1 == root.Val {
				decr = max(decr, right[1]+1)
			}
		}
		ans = max(ans, incr+decr-1)
		return []int{incr, decr}
	}
	dfs(root)
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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