129. Sum Root to Leaf Numbers [LeetCode]

Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path1->2->3which represents the number123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path1->2represents the number12.
The root-to-leaf path1->3represents the number13.

Return the sum = 12 + 13 =25.

笨办法：DFS得出所有路径统一相加

// solution 1
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        List<List<Integer>> paths = new ArrayList<>();
        List<Integer> path = new ArrayList<Integer>();
        dfs(root, path, paths);
        int sum = 0;
        for (List<Integer> way : paths) {
            int mult = 1;
            int curtSum = 0;
            for (int i = way.size() - 1; i >= 0; i--) {
                curtSum += way.get(i) * mult;
                mult *= 10;
            }
            sum += curtSum;
        }
        return sum;
    }

    private void dfs(TreeNode node, List<Integer> path, List<List<Integer>> paths) {
        if (node == null) {
            return;
        }
        path.add(node.val);
        if (node.left == null && node.right == null) {
            paths.add(new ArrayList<Integer>(path));
        }
        dfs(node.left, path, paths);
        dfs(node.right, path, paths);
        path.remove(path.size() - 1);
    }
}

Divide & Conquer：通过参数sum向下传递本节点以上的值是什么，通过return的值向上返回本节点以下的相加和

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        return getSum(root, 0);
    }

    private int getSum(TreeNode node, int sum) {
        if (node == null) {
            return 0;
        }
        if (node.left == null && node.right == null) {
            return sum * 10 + node.val;
        }
        int leftSum = getSum(node.left, sum * 10 + node.val);
        int rightSum = getSum(node.right, sum * 10 + node.val);
        return leftSum + rightSum;
    }
}