05 August 2008
Given a binary tree, return all root-to-leaf paths.
Note: A leaf is a node with no children.
Example:
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Input:
1
/ \
2 3
\
5
Output: ["1->2->5", "1->3"]
Explanation: All root-to-leaf paths are: 1->2->5, 1->3
直接递归来做，比较简单，时间和空间都是O(n)。
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/**
* 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<>();
if (root == null) {
return result;
}
findPaths(root, result, "");
return result;
}
private void findPaths(TreeNode root, List<String> result, String onePath) {
if (root.left == null && root.right == null) {
result.add(onePath + root.val);
}
if (root.left != null) {
findPaths(root.left, result, onePath + root.val + "->");
}
if (root.right != null) {
findPaths(root.right, result, onePath + root.val + "->");
}
}
}
另外一种递归的写法，不用额外的方法
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/**
* 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> paths = new ArrayList<String>();
if (root == null) {
return paths;
}
if (root.left == null && root.right== null) {
paths.add(root.val + "");
return paths;
}
for (String path : binaryTreePaths(root.left)) {
paths.add(root.val + "->" + path);
}
for (String path : binaryTreePaths(root.right)) {
paths.add(root.val + "->" + path);
}
return paths;
}
}