Question

549. Binary Tree Longest Consecutive Sequence II

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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

# 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

/**
 * 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};
  }
}

/**
 * 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};
  }
};

/**
 * 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
}