Problem 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:

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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].

Solution:

If the longest path will include the root node, then the longest path must be the depth(root->right) + depth (root->left)
If the longest path does not include the root node, this problem is divided into 2 sub-problem: set left child and right child as the new root separately, and repeat step1.

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class Solution:
    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        self.max_d = 0
        if not root:
            return 0
        
        def dfs(root):
            if not root:
                return 0
            left = dfs(root.left)
            right = dfs(root.right)
            self.max_d = max(self.max_d, left+right)
            return max(left, right)+1
        dfs(root)
        return self.max_d
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int max_d= 0;
    int diameterOfBinaryTree(TreeNode* root) {
        /*
        design a maxDepth function to calculate depth for left and right subtree
        a global variable for maximum diameter
        */
        if(!root) return 0;
        maxDepth(root);
        return max_d;
        
    }
    
    int maxDepth(TreeNode* root){
        if(!root) return 0;
        int left= 0, right= 0;
        if(root->left) left= maxDepth(root->left);
        if(root->right) right= maxDepth(root->right);
        
        max_d= max(max_d, left+right);
        return max(left, right)+1;
    }
};

time complexity: $O(n)$
space complexity: $O(1)$
reference:
https://goo.gl/PFpQAY