Skip to content

549. Binary Tree Longest Consecutive Sequence II πŸ”’

DifficultyMedium

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

Solution 1

Thinking

A path may bend at a node and may increase or decrease, so it is not a one-way parent-to-child chain. Restarting a search at every node repeats work.

DFS returns the longest increasing and decreasing runs that start at this node and go toward the parent. A child whose value differs by \(1\) extends the matching run. The answer is \(incr+decr-1\) (the node is counted twice). Only one-sided lengths go upward.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
# 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)
                if root.right.val - 1 == root.val:
                    decr = max(decr, d2 + 1)
            ans = max(ans, incr + decr - 1)
            return [incr, decr]

        ans = 0
        dfs(root)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
/**
 * 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};
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans;

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

    vector<int> dfs(TreeNode* root) {
        if (!root) return {0, 0};
        int incr = 1, decr = 1;
        auto left = dfs(root->left);
        auto right = dfs(root->right);
        if (root->left) {
            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) {
            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 {incr, decr};
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
/**
 * 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
}

Comments