Problem:
Given a binary tree, return all root-to-leaf paths.For example, given the following binary tree:
1
/ \
2 3
\
5
All root-to-leaf paths are:
["1->2->5", "1->3"]
Java Code:
/**
* 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> result = new ArrayList<String>();
if (root == null) {
return result;
}
List<Integer> path = new ArrayList<Integer>();
helper(root, path, result);
return result;
}
private void helper(TreeNode root, List<Integer> path, List<String> result) {
path.add(root.val);
if (root.left != null) {
helper(root.left, path, result);
}
if (root.right != null) {
helper(root.right, path, result);
}
if (root.left == null && root.right == null) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < path.size(); i++) {
sb.append(path.get(i));
if (i != path.size() - 1) {
sb.append("->");
}
}
result.add(new String(sb));
}
path.remove(path.size() - 1);
}
}
这个递归解法感觉有些复杂了,有没有更简单的解法呢?
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