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) 求樹的高度，int depthDiff(TreeNode* root)求root的左子樹和右子樹的高度和。遞歸求樹的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));
    }
};