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298. Binary Tree Longest Consecutive Sequence πŸ”’

Description

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

A consecutive sequence path is a path where the values increase by one along the path.

Note that the path can start at any node in the tree, and you cannot go from a node to its parent in the path.

 

Example 1:

Input: root = [1,null,3,2,4,null,null,null,5]
Output: 3
Explanation: Longest consecutive sequence path is 3-4-5, so return 3.

Example 2:

Input: root = [2,null,3,2,null,1]
Output: 2
Explanation: Longest consecutive sequence path is 2-3, not 3-2-1, so return 2.

 

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 consecutive path is a downward chain whose values increase by \(1\). Postorder yields the lengths from each child; we extend only when the child is exactly one larger, otherwise restart at \(1\).

\(dfs\) returns the longest consecutive length starting at the current node and updates a global answer.

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# 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: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l = dfs(root.left) + 1
            r = dfs(root.right) + 1
            if root.left and root.left.val - root.val != 1:
                l = 1
            if root.right and root.right.val - root.val != 1:
                r = 1
            t = max(l, r)
            nonlocal ans
            ans = max(ans, t)
            return t

        ans = 0
        dfs(root)
        return ans

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/**
 * 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) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left) + 1;
        int r = dfs(root.right) + 1;
        if (root.left != null && root.left.val - root.val != 1) {
            l = 1;
        }
        if (root.right != null && root.right.val - root.val != 1) {
            r = 1;
        }
        int t = Math.max(l, r);
        ans = Math.max(ans, t);
        return t;
    }
}

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/**
 * 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 longestConsecutive(TreeNode* root) {
        int ans = 0;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left) + 1;
            int r = dfs(root->right) + 1;
            if (root->left && root->left->val - root->val != 1) {
                l = 1;
            }
            if (root->right && root->right->val - root->val != 1) {
                r = 1;
            }
            int t = max(l, r);
            ans = max(ans, t);
            return t;
        };
        dfs(root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestConsecutive(root *TreeNode) (ans int) {
    var dfs func(*TreeNode) int
    dfs = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        l := dfs(root.Left) + 1
        r := dfs(root.Right) + 1
        if root.Left != nil && root.Left.Val-root.Val != 1 {
            l = 1
        }
        if root.Right != nil && root.Right.Val-root.Val != 1 {
            r = 1
        }
        t := max(l, r)
        ans = max(ans, t)
        return t
    }
    dfs(root)
    return
}

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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function longestConsecutive(root: TreeNode | null): number {
    let ans = 0;
    const dfs = (root: TreeNode | null): number => {
        if (root === null) {
            return 0;
        }
        let l = dfs(root.left) + 1;
        let r = dfs(root.right) + 1;
        if (root.left && root.left.val - root.val !== 1) {
            l = 1;
        }
        if (root.right && root.right.val - root.val !== 1) {
            r = 1;
        }
        const t = Math.max(l, r);
        ans = Math.max(ans, t);
        return t;
    };
    dfs(root);
    return ans;
}

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