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【代码随想录算法训练营第37期第十七天 | LeetCode110.平衡二叉树、257. 二叉树的所有路径、404.左叶子之和】

news 来源：原创 2024/7/7 20:36:13

代码随想录算法训练营第37期第十七天 | LeetCode110.平衡二叉树、257. 二叉树的所有路径、404.左叶子之和

一、110.平衡二叉树

解题代码C++：

/*** 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 getDepth(TreeNode* node){if(!node) return 0;return 1 + max(getDepth(node->left), getDepth(node->right));}bool isBalanced(TreeNode* root) {if(!root) return true;return abs(getDepth(root->left) - getDepth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right);}
};

题目链接/文章讲解/视频讲解：
https://programmercarl.com/0110.%E5%B9%B3%E8%A1%A1%E4%BA%8C%E5%8F%89%E6%A0%91.html

二、257. 二叉树的所有路径

解题代码C++：

/*** 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:void traversal(TreeNode* cur, string path, vector<string>& result){path += to_string(cur->val);if(cur->left == NULL && cur->right == NULL){result.push_back(path);return;}if(cur->left) traversal(cur->left, path + "->", result);if(cur->right) traversal(cur->right, path + "->", result);}vector<string> binaryTreePaths(TreeNode* root) {vector<string> result;string path;if(!root) return result;traversal(root, path, result);return result;}
};

题目链接/文章讲解/视频讲解：
https://programmercarl.com/0257.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%89%80%E6%9C%89%E8%B7%AF%E5%BE%84.html

三、404.左叶子之和

解题代码C++：

/*** 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 sumOfLeftLeaves(TreeNode* root) {if(!root) return 0;int leftValue = 0;if(root->left != NULL && root->left->left == NULL && root->left->right == NULL)leftValue += root->left->val;return leftValue + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);}
};