Leetcode: Sum Root to Leaf Numbers

题目：

Given a binary tree containing digits from

0-9

only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path

1->2->3

which represents the number

123

.

Find the total sum of all root-to-leaf numbers.

For example,

1
/ \
2   3

The root-to-leaf path

1->2

represents the number

12

.

The root-to-leaf path

1->3

represents the number

13

.

Return the sum = 12 + 13 =

25

.

思路分析：

求二叉树根到叶节点的路径和。

采用二叉树的深度优先遍历，遍历过程中记录路径并求和。

C++参考代码：

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
private:
//二叉树深度优先遍历
void dfs(TreeNode *root, int &sum, int num)
{
if (!root->left && !root->right) sum += num;
if (root->left) dfs(root->left, sum, num * 10 + root->left->val);
if (root->right) dfs(root->right, sum, num * 10 + root->right->val);
}

public:
int sumNumbers(TreeNode *root)
{
if (!root) return 0;
int sum = 0;
dfs(root, sum, root->val);
return sum;
}
};
C#参考代码：
/**
* Definition for binary tree
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution
{
private int sum = 0;
private void DFS(TreeNode root, int num)
{
if (root.left == null && root.right == null) sum += num;
if (root.left != null) DFS(root.left, num * 10 + root.left.val);
if (root.right != null) DFS(root.right, num * 10 + root.right.val);
}

public int SumNumbers(TreeNode root)
{
if (root == null) return 0;
DFS(root, root.val);
return sum;
}
}