LeetCode（剑指 Offer）- 37. 序列化二叉树

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题目大意：略

解题思路：略

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AC 代码

• Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));
 
// 解决方案(1)
public class Codec {
 
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
 
        Queue<TreeNode> queue = new LinkedList<TreeNode>(){{offer(root);}};
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node == null) {
                sb.append(Integer.MAX_VALUE).append("#");
                continue;
            }
            sb.append(node.val).append("#");
            queue.offer(node.left);
            queue.offer(node.right);
        }
        return sb.toString();
    }
 
    public TreeNode deserialize(String data) {
        String[] arr = data.split("#");
        int ptrPar = arr.length - 1, ptrSub = arr.length - 1;
 
        Map<Integer, TreeNode> nodeMap = new HashMap<>();
        Map<TreeNode, TreeNode> parLMap = new HashMap<>();
        Map<TreeNode, TreeNode> parRMap = new HashMap<>();
        boolean flag = true;
        while (ptrPar >= 0) {
            // 定位父亲指针
            while (arr[ptrPar].equals(String.valueOf(Integer.MAX_VALUE))) {
                ptrPar--;
            }
 
            // 生成父节点
            TreeNode parNode = new TreeNode(Integer.valueOf(arr[ptrPar]));
            nodeMap.put(ptrPar--, parNode);
 
            // 生成儿子节点
            TreeNode lNode, rNode;
            if (nodeMap.containsKey(ptrSub)) {
                rNode = nodeMap.get(ptrSub);
            } else {
                Integer arrValue = Integer.valueOf(arr[ptrSub]);
                rNode = new TreeNode(arrValue);
                if (!arrValue.equals(Integer.MAX_VALUE)) {
                    flag = false;
                }
                if (flag) {
                    parRMap.put(rNode, parNode);
                }
            }
            ptrSub--;
 
            if (nodeMap.containsKey(ptrSub)) {
                lNode = nodeMap.get(ptrSub);
            } else {
                Integer arrValue = Integer.valueOf(arr[ptrSub]);
                lNode = new TreeNode(arrValue);
                if (!arrValue.equals(Integer.MAX_VALUE)) {
                    flag = false;
                }
                if (flag) {
                    parLMap.put(lNode, parNode);
                }
            }
            ptrSub--;
 
            // 组合节点
            parNode.right = rNode;
            parNode.left = lNode;
        }
 
        // 满二叉树转换完全二叉树(去除末尾的 NULL)
        convertTree(parLMap, parRMap);
 
        return nodeMap.get(0);
    }
 
    private void convertTree(Map<TreeNode, TreeNode> parLMap, Map<TreeNode, TreeNode> parRMap) {
        for (Map.Entry<TreeNode, TreeNode> item : parLMap.entrySet()) {
            item.getValue().left = null;
        }
        for (Map.Entry<TreeNode, TreeNode> item : parRMap.entrySet()) {
            item.getValue().right = null;
        }
    }
}
 
// 解决方案(2)
public class Codec {
    public String serialize(TreeNode root) {
        if(root == null) return "[]";
        StringBuilder res = new StringBuilder("[");
        Queue<TreeNode> queue = new LinkedList<>() {{ add(root); }};
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if(node != null) {
                res.append(node.val + ",");
                queue.add(node.left);
                queue.add(node.right);
            }
            else res.append("null,");
        }
        res.deleteCharAt(res.length() - 1);
        res.append("]");
        return res.toString();
    }
 
    public TreeNode deserialize(String data) {
        if(data.equals("[]")) return null;
        String[] vals = data.substring(1, data.length() - 1).split(",");
        TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
        Queue<TreeNode> queue = new LinkedList<>() {{ add(root); }};
        int i = 1;
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if(!vals[i].equals("null")) {
                node.left = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.left);
            }
            i++;
            if(!vals[i].equals("null")) {
                node.right = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.right);
            }
            i++;
        }
        return root;
    }
}

• C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));
 
class Codec {
public:
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        if(root == nullptr) return "[]";
        string res = "[";
        queue<TreeNode*> que;
        que.emplace(root);
        while(!que.empty()) {
            TreeNode* node = que.front();
            que.pop();
            if(node != nullptr) {
                res += (to_string(node->val) + ",");
                que.emplace(node->left);
                que.emplace(node->right);
            }
            else res += "null,";
        }
        res.pop_back();
        res += "]";
        return res;
    }
 
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if (data == "[]")
            return nullptr;
        vector<string> list = split(data.substr(1, data.length() - 2), ",");
        TreeNode *root = new TreeNode(std::stoi(list[0]));
        queue<TreeNode*> que;
        que.emplace(root);
        int i = 1;
        while(!que.empty()) {
            TreeNode *node = que.front();
            que.pop();
            if(list[i] != "null") {
                node->left = new TreeNode(std::stoi(list[i]));
                que.emplace(node->left);
            }
            i++;
            if(list[i] != "null") {
                node->right = new TreeNode(std::stoi(list[i]));
                que.emplace(node->right);
            }
            i++;
        }
        return root;
    }
    
private:
    // Split a str by a delim
    vector<string> split(string str, string delim) {
        vector<string> list;
        int i = 0, j = 0, len = str.length();
        while (i < len) {
            while (j < len && str[j] != ',')
                j++;
            list.push_back(str.substr(i, j - i));
            i = ++j;
        }
        return list;
    }
};