110.平衡二叉树

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

 使用动态规划，刚开始设置动态规划出来的。

接着求左右节点的高度，当然在求完左右节点之后要确认是否是-1，如果是-1就直接返回了，

如果不是-1就作差值，高度差>1就返回-1，否则得到中间节点的高度。

 感觉动态规划还是很麻烦的。。。

/*** 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 GetBool(TreeNode* Cur) {if (Cur == nullptr) return 0;int left = GetBool(Cur->left); if (left == -1) return -1;int right = GetBool(Cur->right);if (right == -1) return -1;if (abs(left - right) > 1) return -1;else return 1 + max(left, right);}bool isBalanced(TreeNode* root) {return GetBool(root) == -1 ? false : true;}
};

257. 二叉树的所有路径

题目链接：257. 二叉树的所有路径 - 力扣（LeetCode）

/*** 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 traveral (TreeNode* Cur, vector<int>& Path, vector<string>& Result) {Path.push_back(Cur->val);if (Cur->left == nullptr && Cur->right == nullptr) {string String;for (int i = 0; i < Path.size() -1; i++) {String += to_string(Path[i]);String += "->";}String += to_string(Path[Path.size() - 1]);Result.push_back(String);return;}if (Cur->left) {traveral(Cur->left,Path, Result);Path.pop_back();}if (Cur->right) {traveral(Cur->right,Path, Result);Path.pop_back();}}vector<string> binaryTreePaths(TreeNode* root) {vector<int> Path;vector<string> Result;if (root == nullptr) return Result;traveral(root, Path, Result);return Result;}
};

404.左叶子之和

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

/*** 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 == NULL) return 0;if (root->left == NULL && root->right== NULL) return 0;int leftValue = sumOfLeftLeaves(root->left);    // 左if (root->left && !root->left->left && !root->left->right) { // 左子树就是一个左叶子的情况leftValue = root->left->val;}int rightValue = sumOfLeftLeaves(root->right);  // 右int sum = leftValue + rightValue;               // 中return sum;}
};