leetcode:Populating Next Right Pointers in Each Node

一道DFS的题目，题意如下：

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：其实类似于打印所有的路径，只是多了一步求和的步骤
代码如下：

static int sum;
static StringBuffer sb;
public static int sumNumbers(TreeNode root) {
if(root == null)return 0;
sb = new StringBuffer();
sum = 0;
s(root);
return sum;
}
static void s(TreeNode root){

if(root == null){
return;
}else if(root.left == null && root.right == null){
sb.append(root.val);
sum += Integer.parseInt(sb.toString());
sb.deleteCharAt(sb.length()-1);
return ;
}
sb.append(root.val);
s(root.left);
s(root.right);
sb.deleteCharAt(sb.length()-1);
return;
}

一开始比较粗心，没有在非叶子节点return的时候删掉当前值，导致溢出了，所以最后两句很重要~