Java

/**

 Given a binary tree, find 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.

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

 Example 1:
 Input:

      5
     / \
    4   5
   / \   \
 1   1   5
 Output: 2

 Example 2:
 Input:

      1
     / \
    4   5
   / \   \
 4   4   5
 Output: 2


 Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

 */

public class Longest_Univalue_Path {
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {

        int longest = 0;

        public int longestUnivaluePath(TreeNode root) {

            if (root == null) {
                return 0;
            }

            dfs(root);

            return longest;
        }

        private int dfs(TreeNode root) {

            if (root == null) {
                return 0;
            }

            int leftLeaf = dfs(root.left);
            int rightLeaf = dfs(root.right);

            int leftCount = 0;
            int rightCount = 0;

            if (root.left != null && root.left.val == root.val) {
                leftCount = 1 + leftLeaf;
            }
            if (root.right != null && root.right.val == root.val) {
                rightCount = 1 + rightLeaf;
            }

            int total = leftCount + rightCount; // here is edge, not node count
            longest = Math.max(longest, total);

            return Math.max(leftCount, rightCount);
        }
    }
}

// OJ: https://leetcode.com/problems/longest-univalue-path/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int dfs(TreeNode *root) {
        if (!root) return 0;
        int left = dfs(root->left), right = dfs(root->right);
        if (!root->left || root->left->val != root->val) left = 0;
        if (!root->right || root->right->val != root->val) right = 0;
        ans = max(ans, 1 + left + right);
        return 1 + max(left, right);
    }
public:
    int longestUnivaluePath(TreeNode* root) {
        dfs(root);
        return max(0, ans - 1);
    }
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        longest = [0]
        def dfs(root):
            if not root:
                return 0
            left_len, right_len = dfs(root.left), dfs(root.right)
            left = left_len + 1 if root.left and root.left.val == root.val else 0
            right = right_len + 1 if root.right and root.right.val == root.val else 0
            longest[0] = max(longest[0], left + right)
            return max(left, right)
        dfs(root)
        return longest[0]

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int ans;

    public int longestUnivaluePath(TreeNode root) {
        ans = 0;
        arrowLength(root);
        return ans;
    }

    public int arrowLength(TreeNode node) {
        if (node == null)
            return 0;
        int left = arrowLength(node.left);
        int right = arrowLength(node.right);
        int arrowLeft = 0, arrowRight = 0;
        if (node.left != null && node.left.val == node.val)
            arrowLeft += left + 1;
        if (node.right != null && node.right.val == node.val)
            arrowRight += right + 1;
        ans = Math.max(ans, arrowLeft + arrowRight);
        return Math.max(arrowLeft, arrowRight);
    }
}

// OJ: https://leetcode.com/problems/longest-univalue-path/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int dfs(TreeNode *root) {
        if (!root) return 0;
        int left = dfs(root->left), right = dfs(root->right);
        if (!root->left || root->left->val != root->val) left = 0;
        if (!root->right || root->right->val != root->val) right = 0;
        ans = max(ans, 1 + left + right);
        return 1 + max(left, right);
    }
public:
    int longestUnivaluePath(TreeNode* root) {
        dfs(root);
        return max(0, ans - 1);
    }
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        longest = [0]
        def dfs(root):
            if not root:
                return 0
            left_len, right_len = dfs(root.left), dfs(root.right)
            left = left_len + 1 if root.left and root.left.val == root.val else 0
            right = right_len + 1 if root.right and root.right.val == root.val else 0
            longest[0] = max(longest[0], left + right)
            return max(left, right)
        dfs(root)
        return longest[0]

687 - Longest Univalue Path

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