Binary Tree Paths
ID: 257; easy
Solution 1 (Go)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
import "strconv"
func binaryTreePaths(root *TreeNode) []string {
if root == nil {
return []string{}
}
if root.Left == nil && root.Right == nil {
return []string{strconv.Itoa(root.Val)}
}
result := make([]string, 0)
leftPaths := binaryTreePaths(root.Left)
for _,lp := range leftPaths {
result = append(result, strconv.Itoa(root.Val) + "->" + lp)
}
rightPaths := binaryTreePaths(root.Right)
for _,rp := range rightPaths {
result = append(result, strconv.Itoa(root.Val) + "->" + rp)
}
return result
}Solution 2 (Java)
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: the root of the binary tree
* @return: all root-to-leaf paths
*/
public List<String> binaryTreePaths(TreeNode root) {
List<String> result = new ArrayList<>();
if (root == null) return result;
dfs(root, String.valueOf(root.val), result);
return result;
}
private void dfs(TreeNode root, String path, List<String> result) {
if (root == null) return;
if (root.left == null && root.right == null)
result.add(path);
if (root.left != null)
dfs(root.left, path + "->" + root.left.val, result);
if (root.right != null)
dfs(root.right, path + "->" + root.right.val, result);
}
}Notes
This solution computes paths as strings, which is shorter and perhaps more straightforward to understand.
Solution 3 (Java)
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: the root of the binary tree
* @return: all root-to-leaf paths
*/
public List<String> binaryTreePaths(TreeNode root) {
List<String> result = new ArrayList<>();
List<TreeNode> path = new ArrayList<>();
path.add(root);
dfs(root, path, result);
return result;
}
private void dfs(TreeNode root, List<TreeNode> path, List<String> result) {
if (root == null) return;
if (root.left == null && root.right == null) {
String str = "";
for (int i = 0; i < path.size(); i++) {
if (i > 0) str += "->";
str += path.get(i).val;
}
result.add(str);
return;
}
path.add(root.left);
dfs(root.left, path, result);
path.remove(path.size() - 1);
path.add(root.right);
dfs(root.right, path, result);
path.remove(path.size() - 1);
}
}Notes
This solution computes paths as paths, which is a similar approach used in Binary Tree Path Sum.
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