LeetCode543. Diameter of Binary Tree

Description

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].

my program

思路：int depth(TreeNode* root) 求树的高度，int depthDiff(TreeNode* root)求root的左子树和右子树的高度和。递归求树的diameter

class Solution { public: int depth(TreeNode* root) { if (root == NULL) return 0; return max(depth(root->left),depth(root->right))+1; } int depthDiff(TreeNode* root) { if(root == NULL) return 0; return depth(root->left)+depth(root->right); } int diameterOfBinaryTree(TreeNode* root) { if(root == NULL) return 0; return max(depthDiff(root),max(diameterOfBinaryTree(root->left),diameterOfBinaryTree(root->right))); } };

Submission Details 106 / 106 test cases passed. Status: Accepted Runtime: 19 ms Your runtime beats 18.82 % of cpp submissions.

other methods

C++ Solution with DFS

class Solution { public: int maxdiadepth = 0; int dfs(TreeNode* root){ if(root == NULL) return 0; int leftdepth = dfs(root->left); int rightdepth = dfs(root->right); if(leftdepth + rightdepth > maxdiadepth) maxdiadepth = leftdepth + rightdepth; return max(leftdepth +1, rightdepth + 1); } int diameterOfBinaryTree(TreeNode* root) { dfs(root); return maxdiadepth; } };

C++_Recursive_with brief explanation

class Solution { public: int diameterOfBinaryTree(TreeNode* root) { if(root == nullptr) return 0; int res = depth(root->left) + depth(root->right); return max(res, max(diameterOfBinaryTree(root->left), diameterOfBinaryTree(root->right))); } int depth(TreeNode* root){ if(root == nullptr) return 0; return 1 + max(depth(root->left), depth(root->right)); } };