297. Serialize and Deserialize B

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/

本以为需要将 Tree 变成 preorder/postorder + inorder，然后再construct from the two strings. 题解只用了一个String，其思路是 preorder 或者 level order，每个非 null 的node 都有对应的两个 node。

preorder/postorder + inorder 的思路，是不允许 duplicated values。而下面的题解，是允许的。

level order solution

public class Codec {
    private final static String SPLITTER = ", ";
    private final static String NULL_VALUE = "n";
    
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder builder = new StringBuilder();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {            
            TreeNode node = queue.poll();
            if (node == null) {
                builder.append(NULL_VALUE).append(SPLITTER);
            } else {
                queue.offer(node.left);
                queue.offer(node.right);
                builder.append(node.val).append(SPLITTER);
            }
        }
        
        return builder.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.startsWith(NULL_VALUE)) {
            return null;
        }
        
        String[] values = data.split(SPLITTER);
        Queue<TreeNode> queue = new LinkedList<>();
        TreeNode root = new TreeNode(Integer.parseInt(values[0]));
        queue.offer(root);
        
        for (int i = 1; i < values.length; i++) {
            TreeNode node = queue.poll();
            if (!values[i].equals(NULL_VALUE)) {
                node.left = new TreeNode(Integer.parseInt(values[i]));
                queue.offer(node.left);
            }
            i++;
            if (!values[i].equals(NULL_VALUE)) {
                node.right = new TreeNode(Integer.parseInt(values[i]));
                queue.offer(node.right);
            }
        }
        
        return root;
    }
}

preorder solution

from: https://www.jianshu.com/p/4c2a929c31c3

private static final String spliter = ",";
    private static final String NN = "X";

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        preorder(root, sb);
        return sb.toString();
    }

    private void preorder(TreeNode node, StringBuilder sb) {
        if (node == null) {
            sb.append(NN).append(spliter);
            return;
        }
        sb.append(node.val).append(spliter);
        preorder(node.left, sb);
        preorder(node.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        Queue<String> queue = new LinkedList<>();
        queue.addAll(Arrays.asList(data.split(spliter)));
        //如何返回root：递归返回的最终就是root
        return buildTree(queue);
    }

    private TreeNode buildTree(Queue<String> queue) {
        String val = queue.poll();
        if (val.equals(NN)) {
            return null;
        }
        TreeNode node = new TreeNode(Integer.valueOf(val));
        node.left = buildTree(queue);
        node.right = buildTree(queue);

        return node;
    }

顺便把 105. Construct Binary Tree from Preorder and Inorder Traversal 写了一遍：
需要琢磨的一点是，当前node 的 right child 在 preorder 中的寻找方式。举个例子很容易想明白，'inStart - inIndex' 为 left subtree 的 size，那么 'preStart + (inStart - inIndex) + 1' 即为 right subtree 的 root，亦即当前 node 的right child。

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return buildHelper(0, 0, inorder.length - 1, preorder, inorder);
    }
    
    private TreeNode buildHelper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
        if (preStart >= preorder.length || inStart > inEnd) {
            return null;
        }
        
        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = findIndex(root.val, inStart, inEnd, inorder);
        int leftSubTreeSize = inIndex - inStart; 
        root.left = buildHelper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
        root.right = buildHelper(preStart + leftSubTreeSize + 1, inIndex + 1, inEnd, preorder, inorder);
        
        return root;
    }
    
    private int findIndex(int target, int start, int end, int[] array) {
        for (int i = start; i <= end; i++) {
            if (array[i] == target) {
                return i;
            }
        }
        
        return -1;
    }
}