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~