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297. Serialize and Deserialize B

297. Serialize and Deserialize B

作者: Super_Alan | 来源:发表于2018-04-22 01:43 被阅读0次

    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;
        }
    }
    

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