Leetcode 257: Binary Tree Paths

Posted 2020-10-16 Keep walking

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"]

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     // define a new node structure to track path, very useful
12     public class TNode
13     {
14         public TreeNode n;
15         public string path;
16         public TNode(TreeNode node, string p)
17         {
18             n = node;
19             path =p;
20         }
21     }
22     
23     public IList<string> BinaryTreePaths(TreeNode root) {
24         var result = new List<string>();        
25         if (root == null) return result;
26         
27         var stack = new Stack<TNode>();
28         stack.Push(new TNode(root, root.val.ToString() + "->"));
29         
30         while (stack.Count > 0)
31         {
32             var n = stack.Pop();
33             
34             if (n.n.left == null && n.n.right == null)
35             {
36                 result.Add(n.path.Substring(0, n.path.Length - 2));
37             }
38             
39             if (n.n.left != null)
40             {
41                 stack.Push(new TNode(n.n.left, n.path + n.n.left.val.ToString() + "->"));
42             }
43             
44             if (n.n.right != null)
45             {
46                 stack.Push(new TNode(n.n.right, n.path + n.n.right.val.ToString() + "->"));
47             }
48         }
49         
50         return result;
51     }
52 }
53 
54 public class Solution1 {
55     public IList<string> BinaryTreePaths(TreeNode root) {
56         var result = new List<string>();
57         DFS(root, new List<string>(), result);
58         return result;
59     }
60     
61     // if we use IList<string> as path, we need to do backtracking
62     // we can use string directly to avoid doing backtracking
63     private void DFS(TreeNode node, IList<string> path, IList<string> result)
64     {
65         if (node == null) 
66         {
67             return;
68         }
69         
70         if (node.left == null && node.right == null)
71         {
72             var sb = new StringBuilder();
73             
74             for (int i = 0; i < path.Count; i++)
75             {
76                 sb.Append(path[i]);
77                 sb.Append("->");
78             }
79             
80             sb.Append(node.val.ToString());
81             
82             result.Add(sb.ToString());
83             return;
84         }
85         
86         path.Add(node.val.ToString());
87         
88         DFS(node.left, path, result);
89         
90         DFS(node.right, path, result);
91         
92         path.RemoveAt(path.Count - 1); 
93     }
94 }