Given a binary tree, return all root-to-leaf paths.

    Note: A leaf is a node with no children.

    Example:

    1. Input:
    3.    1
    4.  /   \
    5. 2     3
    6.  \
    7.   5
    9. Output: ["1->2->5", "1->3"]
    11. Explanation: All root-to-leaf paths are: 1->2->5, 1->3

    题意

    打印一棵树从根结点到叶结点的所有路径。

    思路

    简单的DFS处理即可。

    代码实现

    1. class Solution {
    2.     public List<String> binaryTreePaths(TreeNode root) {
    3.         List<String> ans = new ArrayList<>();
    5.         if (root != null) {
    6.             dfs(root, ans, "");
    7.         }
    9.         return ans;
    10.     }
    12.     private void dfs(TreeNode x, List<String> ans, String s) {
    13.         s += s.equals("") ? x.val : "->" + x.val;
    15.         if (x.left != null) {
    16.             dfs(x.left, ans, s);
    17.         }
    19.         if (x.right != null) {
    20.             dfs(x.right, ans, s);
    21.         }
    23.         if (x.left == null && x.right == null) {
    24.             ans.add(s);
    25.         }
    26.     }
    27. }