Question
Formatted question description: https://leetcode.ca/all/257.html
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"]
@tag-tree
Algorithm
In the recursive function, when a leaf node is encountered, there is no left or right child node, then a complete path has been formed at this time, add the current leaf node and save it in the result, and then backtrack.
Code
Java
import java.util.ArrayList;
import java.util.List;
public class Binary_Tree_Paths {
public static void main (String[] args) {
Binary_Tree_Paths out = new Binary_Tree_Paths();
Solution s = out.new Solution();
TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.right = new TreeNode(5);
System.out.println(s.binaryTreePaths(root));
Solution_iteration si = out.new Solution_iteration();
System.out.println(si.binaryTreePaths(root));
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution_iteration {
public List<Integer> binaryTreePaths(TreeNode root) {
if (root == null) {
return null;
}
Binary_Tree_Preorder_Traversal preOderIteration = new Binary_Tree_Preorder_Traversal();
Binary_Tree_Preorder_Traversal.Solution preOrderSolution = preOderIteration.new Solution();
List<Integer> result = preOrderSolution.preorderTraversal(root);
return result;
}
}
/**
* 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<String>();
if(root == null) {
return result;
}
helper(new String(), root, result);
return result;
}
public void helper(String current, TreeNode root, List<String> result) {
if(root.left == null && root.right == null) {
result.add(current + root.val);
}
if(root.left != null) {
helper(current + root.val + "->", root.left, result);
}
if(root.right != null) {
helper(current + root.val + "->", root.right, result);
}
}
}
}