Binary Tree Paths 解答

Question

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"]

Solution -- Recursive

We can not apply inorder traversal here because it only show path from root to leaf node as required. Therefore, we use recursive way.

We consider two situation:

1. Current node is leaf. Add string to result.

2. Current node has children. Go on.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<String> binaryTreePaths(TreeNode root) {
12         List<String> result = new ArrayList<String>();
13         if (root == null)
14             return result;
15         dfs(root, result, new StringBuilder());
16         return result;
17     }
18     
19     private void dfs(TreeNode root, List<String> result, StringBuilder current) {
20         if (root == null)
21             return;
22         if (root.left == null && root.right == null) {
23             current.append(root.val);
24             result.add(current.toString());
25             return;
26         }
27         
28         current.append(root.val);
29         current.append("->");
30         
31         StringBuilder tmp1 = new StringBuilder(current);
32         StringBuilder tmp2 = new StringBuilder(current);
33         if (root.left != null)
34             dfs(root.left, result, tmp1);
35         if (root.right != null)
36             dfs(root.right, result, tmp2);
37     }
38 }

 

posted @ 2015-09-30 05:00  树獭君  阅读(184)  评论(0编辑  收藏  举报