leetcode 257 Binary Tree Paths-蒲公英云

leetcode 257 Binary Tree Paths

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

Note: A leaf is a node with no children.

Example:

Input:
   1
 /   \
2     3
 \
  5
Output: ["1->2->5", "1->3"]
Explanation: All root-to-leaf paths are: 1->2->5, 1->3

递归方法：

1、递归终止条件，root是否为空，及是否为叶子结点

2、递归过程

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<String>();
        if(root == null)
            return res;
        if(root.left == null && root.right == null){
            res.add(root.val + "");
            return res;
        }
        List<String> leftS = binaryTreePaths(root.left);
        for(int i=0;i<leftS.size();i++){
            res.add(root.val + "->" + leftS.get(i));
        }
        List<String> rightS = binaryTreePaths(root.right);
        for(int i=0;i<rightS.size();i++){
            res.add(root.val + "->" + rightS.get(i));
        }
        return res;
    }
}

submission中深度优先遍历DFS：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new LinkedList<>();
        dfs(result, root, "");
        return result;
    }
    private void dfs(List<String> result, TreeNode node, String s) {
        if (node == null) return;
        String built = s.length() == 0 ? s + node.val : s + "->" + node.val;
        if (node.left == null && node.right == null) {
            result.add(built);
        } else {
            dfs(result, node.left, built);
            dfs(result, node.right, built);
        }
    }
}