Question

298. Binary Tree Longest Consecutive Sequence

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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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}