From 008a8234cfe728c42659a7007dc8b36907893dee Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Wed, 10 Jul 2019 23:24:28 +0800
Subject: [PATCH] Create 129. Sum Root to Leaf Numbers.md

---
 solutions/129. Sum Root to Leaf Numbers.md | 24 ++++++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 solutions/129. Sum Root to Leaf Numbers.md

diff --git a/solutions/129. Sum Root to Leaf Numbers.md b/solutions/129. Sum Root to Leaf Numbers.md
new file mode 100644
index 0000000..f9e8321
--- /dev/null
+++ b/solutions/129. Sum Root to Leaf Numbers.md	
@@ -0,0 +1,24 @@
+# [129. Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers/)
+# 思路
+采用递归的思路：用sum记录根到某节点组成数字的值，那么每遇到一个子结点的数字，要把之前sum的数字扩大10倍之后再加上当前节点的值。
+如果遍历到叶结点了，就将当前的累加结果sum返回。如果不是，则对其左右子结点分别调用递归函数，将两个结果相加返回即可。
+
+当然也可以改成非递归的，可以用当前节点的val记录上述的sum：将自己的val加上父节点的val乘10作为新的val，例如`node->left->val += 10 * node->val`，代码这里
+就不放了。
+
+# C++
+``` C++
+class Solution {
+private:
+    int helper(TreeNode* root, int sum){
+        if(!root) return 0;
+        sum = sum * 10 + root -> val;
+        if(!root -> left && !root -> right) return sum;
+        return helper(root -> left, sum) + helper(root -> right, sum);
+    }
+public:
+    int sumNumbers(TreeNode* root) {
+        return helper(root, 0);
+    }
+};
+```