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LeetCode OJ Binary Tree Paths

时间:2015-08-28 17:37:44      阅读:168      评论:0      收藏:0      [点我收藏+]

标签:binary tree paths

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"]

递归的方法:

public List<String> binaryTreePaths(TreeNode root) {
    List<String> answer = new ArrayList<String>();
    if (root != null) 
       searchBT(root, "", answer);
    return answer;
}
private void searchBT(TreeNode root, String path, List<String> answer) {
    if (root.left == null && root.right == null) 
        answer.add(path + root.val);//并没有将path的值改变掉
    if (root.left != null) 
        searchBT(root.left, path + root.val + "->", answer);//同理,没有将path的值改变掉,因为后面可能要用到这个值
    if (root.right != null) 
        searchBT(root.right, path + root.val + "->", answer);
}

版权声明:本文为博主原创文章,转载请注明出处:http://blog.csdn.net/lingzhm

LeetCode OJ Binary Tree Paths

标签:binary tree paths

原文地址:http://blog.csdn.net/lingzhm/article/details/48054091

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