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.
Solution
1. Recursive
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int diameterOfBinaryTree(TreeNode* root) {
if (root == nullptr) return 0;
int diameter = getHeight(root->left) + getHeight(root->right);
return max(max(diameterOfBinaryTree(root->left), diameterOfBinaryTree(root->right)), diameter);
}
int getHeight(TreeNode* root) {
if (root == nullptr) return 0;
return max(getHeight(root->left), getHeight(root->right)) + 1;
}
};
2. Optimized
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int diameterOfBinaryTree(TreeNode* root) {
int height = 0;
return diameterOfBinaryTree(root, height);
}
int diameterOfBinaryTree(TreeNode* root, int &height) {
if (root == nullptr) {
height = 0; return 0;
}
int left_height, right_height;
int diameter_left = diameterOfBinaryTree(root->left, left_height);
int diameter_right = diameterOfBinaryTree(root->right, right_height);
height = max(left_height, right_height) + 1;
return max(max(diameter_left, diameter_right), left_height + right_height);
}
};
