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leetcode449. Serialize and Deserialize BST

Honwhy / 3357人閱讀

摘要：題目要求將二叉搜索樹序列化和反序列化，序列化是指將樹用字符串的形式表示，反序列化是指將字符串形式的樹還原成原來的樣子。假如二叉搜索樹的節(jié)點(diǎn)較多，該算法將會占用大量的額外空間。

題目要求 
Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

將二叉搜索樹序列化和反序列化，序列化是指將樹用字符串的形式表示，反序列化是指將字符串形式的樹還原成原來的樣子。

思路和代碼

對于樹的序列化，可以直接聯(lián)想到對樹的遍歷。樹的遍歷包括前序遍歷，中序遍歷，后序遍歷和水平遍歷，并且可知前序遍歷和中序遍歷，或中序遍歷和后序遍歷可以構(gòu)成一棵唯一的樹。除此以外，因?yàn)檫@是一棵二叉搜索樹，可知該樹的中序遍歷就是所有元素的從小到大的排列。

舉個例子，假如一棵樹的結(jié)構(gòu)如下：

  3
 / 
2   4
 
  1

該樹的前序遍歷結(jié)果為3,2,1,4，中序遍歷為1,2,3,4。再仔細(xì)分析前序遍歷的結(jié)果，結(jié)合二叉搜索樹可知，比中間節(jié)點(diǎn)小的值一定位于左子樹，反之一定位于右子樹，即可以對前序遍歷進(jìn)行分割3,|2,1,|4。也就是說，我們可以只利用前序遍歷，就可以區(qū)分出二叉搜索樹的左子樹和右子樹。
代碼如下：

    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        preorder(root, sb);
        return sb.toString();
    }

    public void preorder(TreeNode root, StringBuilder result) {
        if(root != null) {
            result.append(root.val);
            result.append(":");
            preorder(root.left, result);
            preorder(root.right, result);
        }
    }
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data==null || data.isEmpty()) return null;
        String[] preorder = data.split(":");
        String[] inorder = Arrays.copyOf(preorder, preorder.length);
        Arrays.sort(inorder, new Comparator(){

            @Override
            public int compare(String o1, String o2) {
                Integer i1 = Integer.valueOf(o1);
                Integer i2 = Integer.valueOf(o2);
                return i1.compareTo(i2);
            }
            
        });
        
        return build(inorder, preorder, 0, 0, inorder.length);
    }
    
    public TreeNode build(String[] inorder, String[] preorder, int inorderStart, int preorderStart, int length) {
        if(length <= 0) return null;
        TreeNode root = new TreeNode(Integer.valueOf(preorder[preorderStart]));
        for(int i = inorderStart ; i < inorderStart+length ; i++) {
            if(inorder[i].equals(preorder[preorderStart])) {
                root.left = build(inorder, preorder, inorderStart, preorderStart+1, i-inorderStart);
                root.right = build(inorder, preorder, i+1, preorderStart+i-inorderStart+1, inorderStart+length-i-1);
                break;
            }
        }
        
        return root;
    }

這里的代碼是直接使用排序生成了二叉搜索樹的中序遍歷的結(jié)果，并利用先序遍歷和中序遍歷構(gòu)造了一棵二叉搜索樹。假如二叉搜索樹的節(jié)點(diǎn)較多，該算法將會占用大量的額外空間?？梢灾挥孟刃虮闅v作為構(gòu)造樹的輸入，代碼如下：

    public TreeNode deserialize(String data) {
        if (data==null) return null;
        String[] strs = data.split(":");
        Queue q = new LinkedList<>();
        for (String e : strs) {
            q.offer(Integer.parseInt(e));
        }
        return getNode(q);
    }
    
    private TreeNode getNode(Queue q) {
        if (q.isEmpty()) return null;
        TreeNode root = new TreeNode(q.poll());//root (5)
        Queue samllerQueue = new LinkedList<>();
        while (!q.isEmpty() && q.peek() < root.val) {
            samllerQueue.offer(q.poll());
        }
        root.left = getNode(samllerQueue);
        root.right = getNode(q);
        return root;
    }

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