/**
 Given a binary tree, you need to compute the length of the diameter of the tree.
 The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

 Example:
 Given a binary tree
    1
   / \
  2   3
 / \
4   5
 Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

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

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

        int max = 0;

        public int diameterOfBinaryTree(TreeNode root) {
            dfs(root);
            return max;
        }

        private int dfs(TreeNode root) {
            if (root == null) {
                return 0;
            }
            int leftDepth = dfs(root.left);
            int rightDepth = dfs(root.right);

            max = Math.max(max, leftDepth + rightDepth);

            return Math.max(leftDepth, rightDepth) + 1;
        }
    }
}

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

/**
 * 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 diameterOfBinaryTree(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);
        ans = Math.max(ans, left + right);
        return 1 + Math.max(left, right);
    }
}

// OJ: https://leetcode.com/problems/diameter-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int postorder(TreeNode *root) {
        if (!root) return 0;
        int left = postorder(root->left), right = postorder(root->right);
        ans = max(ans, left + right);
        return 1 + max(left, right);
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        postorder(root);
        return ans;
    }
};

# 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 diameterOfBinaryTree(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            nonlocal ans
            left, right = dfs(root.left), dfs(root.right)
            ans = max(ans, left + right)
            return 1 + 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 diameterOfBinaryTree(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """
    self.ans = 0

    def dfs(root):
      if not root:
        return 0
      left = dfs(root.left)
      right = dfs(root.right)
      self.ans = max(self.ans, left + right)
      return max(left, right) + 1

    dfs(root)
    return self.ans

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func diameterOfBinaryTree(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)
		ans = max(ans, left+right)
		return 1 + 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 diameterOfBinaryTree(root: TreeNode | null): number {
    let res = 0;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return 0;
        }
        const { left, right } = root;
        const l = dfs(left);
        const r = dfs(right);
        res = Math.max(res, l + r);
        return Math.max(l, r) + 1;
    };
    dfs(root);
    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>>>, 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, res);
        let right = Self::dfs(&root.right, res);
        *res = (*res).max(left + right);
        left.max(right) + 1
    }

    pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut res = 0;
        Self::dfs(&root, &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 path;

    public int diameterOfBinaryTree(TreeNode root) {
        path = 1;
        depthFirstSearch(root);
        return path - 1;        
    }

    public int depthFirstSearch(TreeNode node) {
        if (node == null)
            return 0;
        int leftDepth = depthFirstSearch(node.left);
        int rightDepth = depthFirstSearch(node.right);
        path = Math.max(path, leftDepth + rightDepth + 1);
        return Math.max(leftDepth, rightDepth) + 1;
    }
}

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

/**
 * 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 diameterOfBinaryTree(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);
        ans = Math.max(ans, left + right);
        return 1 + Math.max(left, right);
    }
}

// OJ: https://leetcode.com/problems/diameter-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int postorder(TreeNode *root) {
        if (!root) return 0;
        int left = postorder(root->left), right = postorder(root->right);
        ans = max(ans, left + right);
        return 1 + max(left, right);
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        postorder(root);
        return ans;
    }
};

# 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 diameterOfBinaryTree(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            nonlocal ans
            left, right = dfs(root.left), dfs(root.right)
            ans = max(ans, left + right)
            return 1 + 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 diameterOfBinaryTree(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """
    self.ans = 0

    def dfs(root):
      if not root:
        return 0
      left = dfs(root.left)
      right = dfs(root.right)
      self.ans = max(self.ans, left + right)
      return max(left, right) + 1

    dfs(root)
    return self.ans

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func diameterOfBinaryTree(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)
		ans = max(ans, left+right)
		return 1 + 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 diameterOfBinaryTree(root: TreeNode | null): number {
    let res = 0;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return 0;
        }
        const { left, right } = root;
        const l = dfs(left);
        const r = dfs(right);
        res = Math.max(res, l + r);
        return Math.max(l, r) + 1;
    };
    dfs(root);
    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>>>, 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, res);
        let right = Self::dfs(&root.right, res);
        *res = (*res).max(left + right);
        left.max(right) + 1
    }

    pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut res = 0;
        Self::dfs(&root, &mut res);
        res
    }
}

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