【问题描述】[简单]

【解答思路】

1. DFS

时间复杂度：O(N^2) 空间复杂度：O(N^2)

class Solution {public List<String> binaryTreePaths(TreeNode root) {List<String> paths = new ArrayList<String>();constructPaths(root, "", paths);return paths;}public void constructPaths(TreeNode root, String path, List<String> paths) {if (root != null) {StringBuffer pathSB = new StringBuffer(path);pathSB.append(Integer.toString(root.val));if (root.left == null && root.right == null) {  // 当前节点是叶子节点paths.add(pathSB.toString());  // 把路径加入到答案中} else {pathSB.append("->");  // 当前节点不是叶子节点，继续递归遍历constructPaths(root.left, pathSB.toString(), paths);constructPaths(root.right, pathSB.toString(), paths);}}}
}

时间复杂度更高 ，使用String+“ ”拼接字符串

class Solution {//这行代码可不敢写这里，leetcode提交会出问题的。。。。!!!//private static List<String> res = new ArrayList<String>();public List<String> binaryTreePaths(TreeNode root) {List<String> res = new ArrayList<String>();if(root == null)return res;BL(root,res,"");return res;}public static void BL(TreeNode node,List<String> res,String str){if(node.left != null) BL(node.left, res, str + node.val + "->");if(node.right != null) BL(node.right, res, str + node.val + "->");if(node.left == null && node.right == null) res.add(str + node.val);}
}链接：https://leetcode-cn.com/problems/binary-tree-paths/solution/chang-gui-cao-zuo-50di-gui-by-gu-xiong-007/

2. BFS

时间复杂度：O(N²) 空间复杂度：O(N²)

class Solution {public List<String> binaryTreePaths(TreeNode root) {List<String> paths = new ArrayList<String>();if (root == null) {return paths;}Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();Queue<String> pathQueue = new LinkedList<String>();nodeQueue.offer(root);pathQueue.offer(Integer.toString(root.val));while (!nodeQueue.isEmpty()) {TreeNode node = nodeQueue.poll(); String path = pathQueue.poll();if (node.left == null && node.right == null) {paths.add(path);} else {if (node.left != null) {nodeQueue.offer(node.left);pathQueue.offer(new StringBuffer(path).append("->").append(node.left.val).toString());}if (node.right != null) {nodeQueue.offer(node.right);pathQueue.offer(new StringBuffer(path).append("->").append(node.right.val).toString());}}}return paths;}
}

【总结】

1. String StringBuffer StringBuilder

2.二叉树遍历

• 前序遍历 先输出当前结点的数据，再依次遍历输出左结点和右结点
• 中序遍历 先遍历输出左结点，再输出当前结点的数据，再遍历输出右结点
• 后续遍历 先遍历输出左结点，再遍历输出右结点，最后输出当前结点的数据

3.复制粘贴时候要小心 修改相应代码 脑子模拟代码 快速排查错误 多从自己写的代码 切忌浮躁

转载链接：https://leetcode-cn.com/problems/binary-tree-paths/solution/er-cha-shu-de-suo-you-lu-jing-by-leetcode-solution/
参考链接：https://leetcode-cn.com/problems/binary-tree-paths/solution/chang-gui-cao-zuo-50di-gui-by-gu-xiong-007/