---------------🎈🎈257. 二叉树的所有路径 题目链接🎈🎈-------------------

解法0 迭代法

解法1 深度优先 前序

时间复杂度O(N)
空间复杂度O(N)

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {List<String> result = new ArrayList<>();public List<String> binaryTreePaths(TreeNode root) {helper(root,"");return result;}public void helper(TreeNode root, String s){// 前序 中左右if(root == null) return;if(root.left == null && root.right == null){result.add(s + root.val);}String temp = s + root.val + "->";helper(root.left,temp);helper(root.right,temp);}
} 

解法2 深度优先 前序 添加了StringBulider

深度优先遍历 添加了StringBulider替代字符串拼接提升效率
toString()转化为String
.append()添加元素

时间复杂度O(N)
空间复杂度O(N)

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {List<String> result = new ArrayList<>();public List<String> binaryTreePaths(TreeNode root) {helper(root,"");return result;}public void helper(TreeNode root, String s){// 前序 中左右  使用StringBuilder提升效率// 停止条件if(root == null) return;if(root.left == null && root.right == null){ // 如果遍历到叶子节点，那么就在result中添加该路径（该路径就是前面得到的s+当前值）result.add(new StringBuilder(s).append(root.val).toString());return;}String temp = new StringBuilder(s).append(root.val).append("->").toString();  // 中 当前路径 s 加上当前节点值和箭头 "->" 的组合helper(root.left, temp);  // 左helper(root.right, temp); // 右}
}