LeetCode 257. Binary Tree Paths

原题链接在这里：https://leetcode.com/problems/binary-tree-paths/

题目：

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"]

题解：

DFS 依次添加，终止条件是叶子节点。

Note:1. 若使用StringBuilder就要做copy.

e.g. [1,2,3], 会写成["1->2","1->23"]. 因为扫过点2，StringBuilder 会变成1->2, 扫过点3，原来的点2没有去掉，就会变成1->23,但是第一个结果没有变成1->23是因为，用到sb.toString()时自动做了copy by value.

2 . 但如果使用String，在递归调用时可以直接使用str, 因为String 和 int, float一样是priority type, 是copy by value，而不像array等object 是 copy by reference.

Time Complexity: O(n), 这是dfs. Space: O(h). h是树的高度，一共用了h层stack.

AC  Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<String>();
        if(root == null){
            return res;
        }
        dfs(root, new StringBuilder(), res);
        return res;
    }
    private void dfs(TreeNode root, StringBuilder sb, List<String> res){
        
        if(root.left == null && root.right == null){
            sb.append(root.val);
            res.add(sb.toString());
            return;
        }
        sb.append(root.val);
        sb.append("->");
        if(root.left != null){
            dfs(root.left,new StringBuilder(sb),res);
        }
        if(root.right != null){
            dfs(root.right,new StringBuilder(sb),res);
        }
    }
}

上面使用StringBuilder, 下面使用String.

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

类似Path Sum, Smallest String Starting From Leaf, Sum Root to Leaf Numbers.