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# 437. Path Sum III (Easy)

https://leetcode.com/problems/path-sum-iii/

/**
 * 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:
    long pathSum(TreeNode *root, int targetSum) {
        return root ? startSumWithRoot(root, targetSum) + pathSum(root->right, targetSum) +
                      pathSum(root->left, targetSum) : 0;
    }
    long startSumWithRoot(TreeNode *root, long targetSum) {
        // 递归终点
        if (!root) {
            return 0;
        }
        long count = 0;
        if(targetSum == root->val) {
            count = 1;
        }
        count += startSumWithRoot(root->left, targetSum - root->val);
        count += startSumWithRoot(root->right, targetSum - root->val);
        return count;
    }
};
# 101. Symmetric Tree (Easy)
https://leetcode.com/problems/symmetric-tree/
/**
 * 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:
    bool isSymmetric(TreeNode* root) {
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode* left, TreeNode* right) {
        if(!left && !right) {
            return true;
        }else if(!left || !right) {
            return false;
        }else{
            if(left->val == right->val) {
                return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
            }else{
                return false;
            }
        }
    }
};