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.
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~leetcode:Populating Next Right Pointers in Each Node
原文地址:http://blog.csdn.net/gotobar/article/details/44930145