129:Sum Root to Leaf Numbers【树】【DFS】

题目链接：click~

/*题意：给出一颗二叉树，从根结点到叶子结点路径上所有结点的值可以组成一个数
求这些数的和
*/

/**
*思路：DFS遍历整棵树，用num记录以当前结点为结尾的可以组成的数
* 当遍历到叶子结点时，累加到sum
*
*/

class Solution {
public:
void GetSum(TreeNode *root, int num, int &sum) {
if(root->left == NULL && root->right == NULL) { //叶子结点
sum += num;
return;
}
if(root->left != NULL)
GetSum(root->left, num*10 + root->left->val, sum);
if(root->right != NULL)
GetSum(root->right, num*10 + root->right->val, sum);

}
int sumNumbers(TreeNode *root) {
if(root == NULL) return 0;
int sum = 0;
GetSum(root, root->val, sum);
return sum;
}
};