Question

687. Longest Univalue Path

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Description

Given the root of a binary tree, return _the length of the longest path, where each node in the path has the same value_. This path may or may not pass through the root.

The length of the path between two nodes is represented by the number of edges between them.

 

Example 1:

Input: root = [5,4,5,1,1,null,5]
Output: 2
Explanation: The shown image shows that the longest path of the same value (i.e. 5).

Example 2:

Input: root = [1,4,5,4,4,null,5]
Output: 2
Explanation: The shown image shows that the longest path of the same value (i.e. 4).

 

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -1000 <= Node.val <= 1000
• The depth of the tree will not exceed 1000.

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 longestUnivaluePath(self, root: TreeNode) -> int:
    def dfs(root):
      if root is None:
        return 0
      left, right = dfs(root.left), dfs(root.right)
      left = left + 1 if root.left and root.left.val == root.val else 0
      right = right + 1 if root.right and root.right.val == root.val else 0
      nonlocal ans
      ans = max(ans, left + right)
      return max(left, right)

    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 longestUnivaluePath(TreeNode root) {
    ans = 0;
    dfs(root);
    return ans;
  }

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

/**
 * 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 longestUnivaluePath(TreeNode* root) {
    ans = 0;
    dfs(root);
    return ans;
  }

  int dfs(TreeNode* root) {
    if (!root) return 0;
    int left = dfs(root->left), right = dfs(root->right);
    left = root->left && root->left->val == root->val ? left + 1 : 0;
    right = root->right && root->right->val == root->val ? right + 1 : 0;
    ans = max(ans, left + right);
    return max(left, right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func longestUnivaluePath(root *TreeNode) int {
  ans := 0
  var dfs func(root *TreeNode) int
  dfs = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    left, right := dfs(root.Left), dfs(root.Right)
    if root.Left != nil && root.Left.Val == root.Val {
      left++
    } else {
      left = 0
    }
    if root.Right != nil && root.Right.Val == root.Val {
      right++
    } else {
      right = 0
    }
    ans = max(ans, left+right)
    return max(left, right)
  }
  dfs(root)
  return ans
}

/**
 * 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 longestUnivaluePath(root: TreeNode | null): number {
  if (root == null) {
    return 0;
  }

  let res = 0;
  const dfs = (root: TreeNode | null, target: number) => {
    if (root == null) {
      return 0;
    }

    const { val, left, right } = root;

    let l = dfs(left, val);
    let r = dfs(right, val);
    res = Math.max(res, l + r);
    if (val === target) {
      return Math.max(l, r) + 1;
    }
    return 0;
  };
  dfs(root, root.val);
  return res;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, target: i32, res: &mut i32) -> i32 {
    if root.is_none() {
      return 0;
    }

    let root = root.as_ref().unwrap().as_ref().borrow();
    let left = Self::dfs(&root.left, root.val, res);
    let right = Self::dfs(&root.right, root.val, res);
    *res = (*res).max(left + right);
    if root.val == target {
      return left.max(right) + 1;
    }
    0
  }

  pub fn longest_univalue_path(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    if root.is_none() {
      return 0;
    }

    let mut res = 0;
    Self::dfs(&root, root.as_ref().unwrap().as_ref().borrow().val, &mut res);
    res
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var longestUnivaluePath = function (root) {
  let ans = 0;
  let dfs = function (root) {
    if (!root) {
      return 0;
    }
    let left = dfs(root.left),
      right = dfs(root.right);
    left = root.left?.val == root.val ? left + 1 : 0;
    right = root.right?.val == root.val ? right + 1 : 0;
    ans = Math.max(ans, left + right);
    return Math.max(left, right);
  };
  dfs(root);
  return ans;
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */

#define max(a, b) (((a) > (b)) ? (a) : (b))

int dfs(struct TreeNode* root, int target, int* res) {
  if (!root) {
    return 0;
  }
  int left = dfs(root->left, root->val, res);
  int right = dfs(root->right, root->val, res);
  *res = max(*res, left + right);
  if (root->val == target) {
    return max(left, right) + 1;
  }
  return 0;
}

int longestUnivaluePath(struct TreeNode* root) {
  if (!root) {
    return 0;
  }
  int res = 0;
  dfs(root, root->val, &res);
  return res;
}