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Java

  • /**
     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;
            }
        }
    }
    
  • // 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
    
    

Java

  • /**
     * 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;
        }
    }
    
  • // 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
    
    

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