257. Binary Tree Paths

Question

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Thinking:

  • Method 1:dfs
/**
 * 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> res = new ArrayList<>();
        if(root == null) return res;
        bfs(root, new StringBuilder(), res);
        return res;
    }
    private void bfs(TreeNode node, StringBuilder sb, List<String> res){
        if(node.left == null && node.right == null){
            sb.append(node.val);
            res.add(sb.toString());
            return;
        }else{
            sb.append(node.val + "->");
            if(node.left != null)   bfs(node.left, new StringBuilder(sb.toString()), res);
            if(node.right != null)   bfs(node.right, new StringBuilder(sb.toString()), res);
        }
    }
}

Second time

  1. I still used bfs to solve this question. However, using list instead of StringBuilder is faster. 
    /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
 private String arrow = "->";
 public List<String> binaryTreePaths(TreeNode root) {
     List<String> result = new LinkedList<>();
     if(root == null) return result;
     addPath(result, new LinkedList<Integer>(), root);
     return result;
 }
 private void addPath(List<String> result, List<Integer> temp, TreeNode node){
     temp.add(node.val);
     if(node.left == null && node.right == null){
         if(temp.size() == 0) return;
         else if(temp.size() == 1){
             result.add("" + temp.get(0));
         }else{
             StringBuilder sb = new StringBuilder();
             sb.append(temp.get(0));
             for(int i = 1; i < temp.size(); i++){
                 sb.append(this.arrow);
                 sb.append(temp.get(i));
             }
             result.add(sb.toString());
         }
     }else{
         if(node.left != null){
             addPath(result, temp, node.left);
         }
         if(node.right != null){
             addPath(result, temp, node.right);
         }
     }
     temp.remove(temp.size() - 1);
 }
}