Skip to main content

543. Diameter of Binary Tree

Given the root of a binary tree, return 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.

The length of a path between two nodes is represented by the number of edges between them.

Example 1:

image

Input: root = [1,2,3,4,5] Output: 3 Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].

Example 2:

Input: root = [1,2] Output: 1

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -100 <= Node.val <= 100

算 Binary Tree 的直徑,dfs 解

/**
* 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 {
private:
int diameter = 0;
public:
int diameterOfBinaryTree(TreeNode* root) {
dfs(root);
return diameter;
}

int dfs(TreeNode* root) {
if(!root) return 0;
// 走訪左邊的 tree
int left = dfs(root->left);

// 走訪右邊的 tree
int right = dfs(root->right);

// 算最大的直徑
diameter = max(diameter, left + right);

// 每次遞迴,要記得直徑 + 1
return 1 + max(left, right);
}
};
  • T: O(N)O(N)
  • S: O(N)O(N)