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257.给Java解法添加更全面清晰的注释,另修改一处重复的笔误
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@ -34,7 +34,7 @@
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## 递归
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1. 递归函数函数参数以及返回值
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1. 递归函数参数以及返回值
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要传入根节点,记录每一条路径的path,和存放结果集的result,这里递归不需要返回值,代码如下:
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@ -259,7 +259,7 @@ if (cur->right) {
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path.pop_back(); // 回溯 '>'
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path.pop_back(); // 回溯 '-'
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}
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```
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```
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整体代码如下:
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@ -395,33 +395,34 @@ class Solution {
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* 递归法
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*/
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public List<String> binaryTreePaths(TreeNode root) {
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List<String> res = new ArrayList<>();
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List<String> res = new ArrayList<>();// 存最终的结果
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if (root == null) {
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return res;
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}
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List<Integer> paths = new ArrayList<>();
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List<Integer> paths = new ArrayList<>();// 作为结果中的路径
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traversal(root, paths, res);
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return res;
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}
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private void traversal(TreeNode root, List<Integer> paths, List<String> res) {
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paths.add(root.val);
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// 叶子结点
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paths.add(root.val);// 前序遍历,中
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// 遇到叶子结点
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if (root.left == null && root.right == null) {
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// 输出
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StringBuilder sb = new StringBuilder();
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StringBuilder sb = new StringBuilder();// StringBuilder用来拼接字符串,速度更快
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for (int i = 0; i < paths.size() - 1; i++) {
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sb.append(paths.get(i)).append("->");
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}
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sb.append(paths.get(paths.size() - 1));
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res.add(sb.toString());
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sb.append(paths.get(paths.size() - 1));// 记录最后一个节点
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res.add(sb.toString());// 收集一个路径
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return;
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}
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if (root.left != null) {
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// 递归和回溯是同时进行,所以要放在同一个花括号里
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if (root.left != null) { // 左
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traversal(root.left, paths, res);
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paths.remove(paths.size() - 1);// 回溯
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}
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if (root.right != null) {
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if (root.right != null) { // 右
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traversal(root.right, paths, res);
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paths.remove(paths.size() - 1);// 回溯
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}
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@ -794,7 +795,7 @@ object Solution {
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}
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}
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```
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rust:
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```rust
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@ -826,3 +827,4 @@ impl Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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