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May 24, 2016 05:37
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Leetcode 297 - Serialize and deserialize binary tree - make the code more readable, run through the test case and then understand the code, also ensure code works.
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using System; | |
using System.Collections.Generic; | |
using System.Linq; | |
using System.Text; | |
using System.Threading.Tasks; | |
namespace _297SerializeDeserializeTree_Stack | |
{ | |
public class TreeNode | |
{ | |
public int val; | |
public TreeNode left; | |
public TreeNode right; | |
public TreeNode(int x) { val = x; } | |
} | |
class Program | |
{ | |
static void Main(string[] args) | |
{ | |
TreeNode node1 = new TreeNode(9); | |
node1.left = new TreeNode(5); | |
node1.right = new TreeNode(8); | |
node1.left.left = new TreeNode(1); | |
node1.left.right = new TreeNode(4); | |
node1.left.right.left = new TreeNode(2); | |
node1.left.right.right = new TreeNode(3); | |
node1.right.right = new TreeNode(7); | |
node1.right.right.left = new TreeNode(6); | |
string result = serialize(node1); | |
TreeNode root = deserialize(result); | |
} | |
/* | |
* May 23, 2016 | |
* Preorder traversal - Use test case to help understand the code quickly. | |
* | |
* ideas to implement the solution: | |
* 1. First, use an array to store a tree node; | |
array[0] = value; | |
array[1] = li, // left child index, if null, -1 | |
array[2] = ri, // right child index, if null, -1 | |
* | |
* 2. nodes will be added to an arrayList of arrays; then, convert the arraylist to a string. | |
* | |
* Very good practice, get familiar with preorder traversal, and how to use it to solve problems. | |
* | |
*/ | |
public static string serialize(TreeNode root) | |
{ | |
string emptyString = ""; | |
if (root == null) return emptyString; | |
List<int[]> nodes = new List<int[]>(); | |
Stack<TreeNode> stack = new Stack<TreeNode>(); | |
Stack<Int32> positionStack = new Stack<Int32>(); | |
TreeNode runner = root; | |
int index = 0; | |
while (true) | |
{ | |
// preorder traversal to construct the arraylist | |
while (runner != null) | |
{ | |
if (stack.Count > 0) | |
{ | |
// get the parent, and set the left and right of the parent. | |
int pi = positionStack.Peek(); // ci is the current index | |
if (runner == stack.Peek().left) | |
{ | |
// cur is left child of the parent | |
nodes[pi][1] = index; | |
} | |
else | |
{ | |
// cur is the right child of the parent | |
nodes[pi][2] = index; | |
} | |
} | |
nodes.Add(new int[] { runner.val, -1, -1 }); | |
stack.Push(runner); | |
positionStack.Push(index); | |
runner = runner.left; | |
++index; | |
} | |
while (stack.Count > 0 && (stack.Peek().right == null || stack.Peek().right == runner)) | |
{ | |
runner = stack.Pop(); | |
positionStack.Pop(); | |
} | |
if (stack.Count == 0 ) | |
break; | |
else | |
runner = stack.Peek().right; | |
} | |
StringBuilder sb = new StringBuilder(); | |
for (int i = 0; i < nodes.Count; ++i) | |
{ | |
for (int j = 0; j < 3; ++j) | |
{ | |
sb.Append(nodes[i][j]); | |
sb.Append(' '); | |
} | |
} | |
return sb.ToString(); | |
} | |
// Decodes your encoded data to tree. | |
public static TreeNode deserialize(string data) | |
{ | |
// convert the string into an arrayList; then, reconstruct the tree using the arraylist. | |
if (data.Length == 0) | |
return null; | |
List<Int32> nodes = new List<Int32>(); | |
int index = 0; | |
while (index < data.Length) | |
{ | |
int start = index; | |
while (data[index] != ' ') | |
++index; | |
int end = index; | |
nodes.Add(Int32.Parse(data.Substring(start, end-start+1))); | |
++index; | |
} | |
List<int[]> list = new List<int[]>(); | |
for (int i = 0; i < nodes.Count / 3; ++i) | |
{ | |
int[] node = new int[3]; | |
for (int j = 0; j < 3; ++j) | |
{ | |
node[j] = nodes[i * 3 + j]; | |
} | |
list.Add(node); | |
} | |
// reconstruct the tree using the list of arrays. | |
TreeNode root = generateTree(list, 0); | |
return root; | |
} | |
private static TreeNode generateTree(List<int[]> list, int index) | |
{ | |
int[] arr = new int[3] { list[index][0], list[index][1], list[index][2] }; | |
TreeNode node = new TreeNode(arr[0]); | |
node.left = arr[1] == -1 ? null : generateTree(list, arr[1]); | |
node.right = arr[2] == -1 ? null : generateTree(list, arr[2]); | |
return node; | |
} | |
} | |
} |
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