/**

 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);
        }
    }
}

############

/**
 * 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);
    }
}

// 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:
#     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.
# 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]

/**
 * 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

/**
 * 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.
 * 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.
// #[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.
 * 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);
    }
}

############

/**
 * 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);
    }
}

// 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:
#     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.
# 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]

/**
 * 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

/**
 * 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.
 * 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.
// #[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
    }
}

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