LeetCode257. Binary Tree Paths

257. Binary Tree Paths

   1 /   \2     3 \  5

["1->2->5", "1->3"]

题意：
给出一个二叉树，返回所有根节点到叶子的路径

public class LeetCode257 {    public List<String> binaryTreePaths(TreeNode root) {        List<String> path = new ArrayList<>();        if(root == null) return path;        String onepath = "";        paths(root, onepath, path);        return path;    }    public static void paths(TreeNode root, String curPath, List<String> path){        curPath = curPath +  root.val ;        if(root.left == null && root.right == null){    // 到达叶子节点            path.add(curPath);        }        curPath = curPath + "->";        if(root.left != null){            paths(root.left, curPath, path);        }        if(root.right != null){            paths(root.right, curPath, path);        }    }}

129. Sum Root to Leaf Numbers
题意：
找到所有根到叶子的数字和
例如：

   1 /   \2     3

1->2 代表12
1->3 代表13
所有根到叶子的和为：12+13=25
解题思路：
思路1：和上题类似，先求出所有路径，推进一步乘以10，求出所有数字，然后相加。
思路2：递归相加

public class LeetCode129 {    /*    public int sumNumbers(TreeNode root) {        List<Integer> path = new ArrayList<>();        int res = 0;        if(root == null) return res;        paths(root, 0, path);        for(Integer tmp:path){            res = res + tmp;            System.out.println(tmp);        }        return res;    }    public static void paths(TreeNode root, int curPath, List<Integer> path){        curPath = curPath +  root.val ;        if(root.left == null && root.right == null){            path.add(curPath);        }        curPath = curPath * 10;        if(root.left != null){            paths(root.left, curPath, path);        }        if(root.right != null){            paths(root.right, curPath, path);        }    }    */    public int sumNumbers(TreeNode root) {          return dfs(root, 0);      }      public int dfs(TreeNode root, int sum) {          if(root == null) return 0;          if(root.left == null && root.right == null) return root.val + sum * 10;                   return dfs(root.left, root.val + sum * 10) + dfs(root.right, root.val + sum * 10);            } }