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