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  • /**
     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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