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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.
My initial solution used an arrayList to hold all root to leaf numbers, and then finally add them all.
public class Solution { ArrayList<Integer> result = new ArrayList<Integer>(); public int sumNumbers(TreeNode root) { if (root == null) return 0; sumNumbersHelper(root, 0); int total = 0; for (int a : result){ total += a; } return total; } public void sumNumbersHelper(TreeNode root, int sum){ if (root == null) return; sum = sum * 10 + root.val; if (root.left == null && root.right == null){ result.add(sum); return; } else { if (root.left != null) sumNumbersHelper(root.left, sum); if (root.right != null) sumNumbersHelper(root.right, sum); } } }
But later I found this much cleaner solution, do not need an arrayList. Instead, just sum them directly.
public class Solution { int result = 0; public int sumNumbers(TreeNode root) { if (root == null) return 0; sumNumbersHelper(root, 0); return result; } public void sumNumbersHelper(TreeNode root, int sum){ if (root == null) return; sum = sum * 10 + root.val; if (root.left == null && root.right == null){ result += sum; return; } else { if (root.left != null) sumNumbersHelper(root.left, sum); if (root.right != null) sumNumbersHelper(root.right, sum); } } }
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原文地址：http://www.cnblogs.com/timoBlog/p/4644486.html
