leetcode 257. Binary Tree Paths（二叉树路径）-优快云博客

本文链接：https://blog.youkuaiyun.com/level_code/article/details/103152980

本文探讨了如何在二叉树中寻找所有从根节点到叶子节点的路径，并使用深度优先搜索(DFS)策略实现这一目标。通过递归方法，文章详细解释了如何将路径保存为字符串，并对比了使用String和StringBuilder的不同效果。

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

Note: A leaf is a node with no children.

Example:

Input:

Output: [“1->2->5”, “1->3”]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

思路：
DFS，对于路径的保存用String，而不是StringBuilder，因为StringBuilder是引用参数，在Parent节点的路径到了child节点做了修改再跳出来的时候，StringBuilder的内容无法回到之前保存的路径
如下面的例子（验证String与StringBuilder，与此题无关）：

	public static void main(String[] args){
		String tmpS = "1";
		StringBuilder tmpB = new StringBuilder();
		tmpB.append(1);
		addString(tmpS);
		addBuilder(tmpB);
		
		 
		System.out.println("tmpS = " + tmpS);
		System.out.println("tmpB = " + tmpB);
	 }
	 
	 public static void addString(String tmp) {
		 tmp += "2";
		 System.out.println("tmpS inside = " + tmp);
	 }
	 
	 public static void addBuilder(StringBuilder tmp1) {
		 tmp1.append("2");
	 }

output:

tmpS inside = 12
tmpS = 1
tmpB = 12

可见String在其他函数做修改跳出后不变，而StringBuilder的值发生了改变

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 ArrayList<>();
        String tmp = "";
        dfs(result, tmp, root);
        
        return result;
    }
    
    public void dfs(List<String> result, String tmp, TreeNode root) {
        if (root == null) {
            return;
        }
        
        tmp += root.val;
        
        if (root.left == null && root.right == null) {
            result.add(tmp);
            return;
        } else {
            tmp += "->";
        }
        
        dfs(result, tmp, root.left);
        dfs(result, tmp, root.right);
    }
}