【Leetcode】Sum Root to Leaf Numbers

题目链接：https://leetcode.com/problems/sum-root-to-leaf-numbers/

题目：

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.

思路：

1、参考Binary Tree Paths 中保存所有从根到叶路径的方法，对所有路径处理一下再相加就能得到结果。这种方法是思路简单，利用 可以保存从根到叶所有路径的方法可以解决很多类似的root-leaf问题。不过需要O(n)的空间保存所有路径。

2、在递归过程中处理路径和。优点是算法简洁，不需要保存路径，空间复杂度O(1)。

算法1：

	List<String> strList = new ArrayList<String>();

	public int sumNumbers(TreeNode root) {
		int sum = 0;
		if (root != null) {
			findPath(root, String.valueOf(root.val));
			for (int i = 0; i < strList.size(); i++) {
				sum += getConbinaFromStr(strList.get(i));
			}
		}
		return sum;
	}

	public void findPath(TreeNode p, String path) {
		if (p.left == null && p.right == null) {
			strList.add(path);
		}
		if (p.left != null)
			findPath(p.left, path + "->" + p.left.val);
		if (p.right != null)
			findPath(p.right, path + "->" + p.right.val);
	}

	public int getConbinaFromStr(String str) {
		int sum = 0;
		String s[] = str.split("->");
		for (int i = 0; i < s.length; i++) {
			int tmp = Integer.parseInt(s[i]);
			sum = sum * 10 + tmp;
		}
		return sum;
	}

算法2：

	int sum = 0;

	public int sumNumbers2(TreeNode root) {
		addPSum(root, 0);
		return sum;
	}

	public void addPSum(TreeNode root, int curNum) {
		if (root == null)
			return;
		curNum = curNum * 10 + root.val; // 此时该路径的和
		if (root.left == null && root.right == null) { // 从根走到叶了
			sum += curNum;
			return;
		}
		if (root.left != null) {
			addPSum(root.left, curNum);
		}
		if (root.right != null) {
			addPSum(root.right, curNum);
		}
	}