LeetCode OJ：Sum Root to Leaf Numbers（根到叶节点数字之和）-优快云博客

本文介绍了一种使用递归算法解决二叉树所有根到叶子节点路径数值之和的方法。通过C++与Java代码实现，简洁高效地计算出所有路径所代表的数字总和。

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.

简单的用递归就可以实现，代码如下：

 1 class Solution {
 2 public:
 3     int sumNumbers(TreeNode* root) {
 4         dfs(root, 0);
 5     }
 6 
 7     int dfs(TreeNode * root, int curr){
 8         if(!root)
 9             return 0;
10         if(!root->left && !root->right)
11             return curr * 10 + root->val;
12         return dfs(root->left, curr    * 10 + root->val) + dfs(root->right, curr * 10 + root->val);
13     }
14 };

java版本的代码如下所示：

 1 public class Solution {
 2     public int sumNumbers(TreeNode root) {
 3         return dfs(root, 0);
 4     }
 5 
 6     public int dfs(TreeNode root, int curr){
 7         if(root == null)
 8             return 0;
 9         if(root.left == null && root.right == null)
10             return curr * 10 + root.val;
11         return dfs(root.left, curr * 10 + root.val) + dfs(root.right, curr * 10 + root.val);
12     }
13 }