LeetCode -- Sum Root to Leaf NNumbers

Related Links:

Path Sum: http://www.cnblogs.com/little-YTMM/p/4529982.html

Path Sum II: http://www.cnblogs.com/little-YTMM/p/4531115.html

Question:

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.

Analysis:

给出一棵二叉树，其节点的值仅在0-9之间，二叉树的每条根节点到叶节点的路径都会产生一个数。

例如一条根-叶的路径1->2->3会产生数字123.

找出所有路径产生数字的和。

思路；由于以前做过path sum的题目，因此用类似的思路，用两个链表分别保存当前节点和当前已经产生的路径值，然后到叶节点时将该路径的值加到最终结果上即可。

Answer：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null)
            return 0;
        LinkedList<TreeNode> node = new LinkedList<TreeNode>();
        LinkedList<Integer> product = new LinkedList<Integer>();
        node.add(root);
        product.add(root.val);
        int result = 0;
        while(!node.isEmpty()) {
            TreeNode tempNode = node.poll();
            int tempProduct = product.poll();
            if(tempNode.left == null && tempNode.right == null)
                result += tempProduct;
            if(tempNode.left != null) {
                node.add(tempNode.left);
                product.add(tempProduct * 10 + tempNode.left.val);
            }
            if(tempNode.right != null) {
                node.add(tempNode.right);
                product.add(tempProduct * 10 + tempNode.right.val);
            }
        }
        return result;
    }
}