257. Binary Tree Paths

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"]

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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Tree Depth-first Search

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(M) Path Sum II

BFS Solution:

public List<String> binaryTreePaths(TreeNode root) {
List<String> answer = new ArrayList<String>();
if (root != null) searchBT(root, "", answer);
return answer;
}
private void searchBT(TreeNode root, String path, List<String> answer) {
if (root.left == null && root.right == null) answer.add(path + root.val);
if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
}

DFS Solution:

/**
* 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) {
ArrayList<String> paths = new ArrayList<String>();
dfs(root, new ArrayDeque<Integer>(), paths);
return paths;
}

private void dfs(TreeNode node, Deque<Integer> path, List<String> paths) {
if(node == null)
return;

path.add(node.val);

if(node.left == null && node.right == null) {
paths.add(mkPath(path));
}

dfs(node.left, path, paths);
dfs(node.right, path, paths);
path.removeLast();
}

private String mkPath(Deque<Integer> path) {
if(path.isEmpty())
return "";

StringBuilder sb  = new StringBuilder();
for(Integer i : path) {
sb.append("->").append(i);
}
return sb.toString().substring(2);
}

}