文章目录 【LC刷题】DAY13：110 257 440 110. 平衡二叉树 [link](https://leetcode.cn/problems/balanced-binary-tree/description/) 257. 二叉树的所有路径 [link](https://leetcode.cn/problems/binary-tree-paths/description/) 404. 左叶子之和 [link](https://leetcode.cn/problems/sum-of-left-leaves/) 222. 完全二叉树的节点个数 [link](https://leetcode.cn/problems/count-complete-tree-nodes/description/)
110. 平衡二叉树 link
/*** 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 getHeight(TreeNode* root){if(root == nullptr){return 0;}int leftHeight = getHeight(root->left);if(leftHeight == -1) return -1;int rightHeight = getHeight(root->right);if (rightHeight == -1) return -1;return abs(leftHeight - rightHeight) > 1 ? -1: 1 + max(leftHeight, rightHeight);}bool isBalanced(TreeNode* root) {return getHeight(root) == -1 ? false : true;}
};
257. 二叉树的所有路径 link
/*** 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 construct_path(TreeNode* node,string path, vector<string>& paths){if(node != nullptr){path += to_string(node->val);if(node->left == nullptr && node -> right == nullptr){paths.push_back(path);}else{path+="->";construct_path(node->left, path, paths);construct_path(node->right, path, paths);}}}vector<string> binaryTreePaths(TreeNode* root) {vector<string> paths;construct_path(root, "", paths);return paths;}
};404. 左叶子之和 link
/*** 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 == nullptr) return 0;if(root->left == nullptr && root->right==nullptr) return 0;int leftValue = sumOfLeftLeaves(root->left);if(root->left && !root->left->left && !root->left->right ){leftValue = root->left->val;}int rightValue = sumOfLeftLeaves(root->right);int sum = leftValue + rightValue;return sum;}
};
222. 完全二叉树的节点个数 link
/*** 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 countNodes(TreeNode* root) {if(root == nullptr) return 0;return 1+countNodes(root->left) + countNodes(root->right);}
};
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