ARST打卡第224周[224/521]

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/**
 * 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 checkSubTree(TreeNode* root, int& sub_sum) {
        // 检查子树是否满足，并且返回子树的sum
        // 递归终止条件是 null
        if (!root) {
            return true;
        }
        // 递归是左右子树
        int l = -101;
        int r = -101;
        if (checkSubTree(root->left, l) && 
            checkSubTree(root->right, r)) {
            // 依次判断是否有左右子树的情况。
            if (l == -101 && r == -101) {
                sub_sum = root->val;
                return true;
            } else if (l == -101) {
                sub_sum = root->val + r;
                return root->val == r;
            } else if (r == -101) {
                sub_sum = root->val + l;
                return root->val == l;
            }
            sub_sum = root->val + l + r;
            return root->val == l + r;
        }
        return false;
    }
    bool checkTree(TreeNode* root) {
        // 简单做法
        // return root->val == root->left->val + root->right->val; 
        // 递归做法
        int root_sum;
        return checkSubTree(root, root_sum);
    }
};