问题：

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

For example, given the following binary tree:

   1 /   \2     3 \  5

All root-to-leaf paths are:

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

解决：

① 简单的二叉树遍历，遍历的过程中记录之前的路径，一旦遍历到叶子节点便将该路径加入结果中。

/**

 * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution { //17ms     List<String> res = new ArrayList<>();     public List<String> binaryTreePaths(TreeNode root) {         if(root != null) findPath(root,String.valueOf(root.val));         return res;     }     public void findPath(TreeNode node,String path){         if(node.left == null && node.right == null) res.add(path);         if(node.left != null) findPath(node.left,path + "->" + node.left.val);         if(node.right != null) findPath(node.right,path + "->" + node.right.val);     } }

② 进化版

public class Solution { //15ms

    public List<String> binaryTreePaths(TreeNode root) {         List<String> res = new ArrayList<String>();         if (root == null) return res;         getPath(root, "", res);         return res;     }     public void getPath(TreeNode root, String path, List<String> res) {         if (root.left == null && root.right == null)             res.add(path + root.val);         else {             if (root.left != null) getPath(root.left, path + root.val + "->", res);             if (root.right != null) getPath(root.right, path + root.val + "->", res);         }     } }

【注意】return影响效率

public class Solution { //15ms

    public List<String> binaryTreePaths(TreeNode root) {         List<String> res = new ArrayList<String>();         if (root == null) return res;         getPath(root, "", res);         return res;     }     public void getPath(TreeNode root, String path, List<String> res) {         if (root.left == null && root.right == null){             res.add(path + root.val);             //return; 加上return耗时为18ms         }         if (root.left != null) getPath(root.left, path + root.val + "->", res);         if (root.right != null) getPath(root.right, path + root.val + "->", res);     } }