449. 序列化和反序列化二叉搜索树

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力扣刷题笔记

Python3版代码提示：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# Your Codec object will be instantiated and called as such:
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# tree = ser.serialize(root)
# ans = deser.deserialize(tree)
# return ans

Java版代码提示：

/**
 * 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();
// String tree = ser.serialize(root);
// TreeNode ans = deser.deserialize(tree);
// return ans;

我真的忘的一干二净，就是有那么一点点熟悉，但是写起来又一脸懵逼。
直接抄题解。

1. 后序遍历

seriallize就是常规的后序遍历（左→右→根）。
deserialize根据二叉搜索树的特性，递归反序列化根节点→右子树（直至接下来需要反序列化的节点值<根节点值）→左子树（直至接下来需要反序列化的节点值>根节点值）

时间复杂度：$O(n)$
空间复杂度：$O(n)$

Python3版：

class Codec:
    def serialize(self, root: TreeNode) -> str:
        arr = []
        def postOrder(root: TreeNode) -> None:
            if root is None:
                return
            postOrder(root.left)
            postOrder(root.right)
            arr.append(root.val)
        postOrder(root)
        return ' '.join(map(str, arr))

    def deserialize(self, data: str) -> TreeNode:
        arr = list(map(int, data.split()))
        def construct(lower: int, upper: int) -> TreeNode:
            if arr == [] or arr[-1] < lower or arr[-1] > upper:
                return None
            val = arr.pop()
            root = TreeNode(val)
            root.right = construct(val, upper)
            root.left = construct(lower, val)
            return root
        return construct(-inf, inf)

Java版：

public class Codec {
    public String serialize(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        postOrder(root, list);
        String str = list.toString();
        return str.substring(1, str.length() - 1);
    }

    public TreeNode deserialize(String data) {
        if (data.isEmpty()) {
            return null;
        }
        String[] arr = data.split(", ");
        Deque<Integer> stack = new ArrayDeque<Integer>();
        int length = arr.length;
        for (int i = 0; i < length; i++) {
            stack.push(Integer.parseInt(arr[i]));
        }
        return construct(Integer.MIN_VALUE, Integer.MAX_VALUE, stack);
    }

    private void postOrder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        postOrder(root.left, list);
        postOrder(root.right, list);
        list.add(root.val);
    }

    private TreeNode construct(int lower, int upper, Deque<Integer> stack) {
        if (stack.isEmpty() || stack.peek() < lower || stack.peek() > upper) {
            return null;
        }
        int val = stack.pop();
        TreeNode root = new TreeNode(val);
        root.right = construct(val, upper, stack);
        root.left = construct(lower, val, stack);
        return root;
    }
}

2. 先序遍历

class Codec:
    def serialize(self, root: Optional[TreeNode]) -> str:
        """Encodes a tree to a single string."""

        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            nums.append(root.val)
            dfs(root.left)
            dfs(root.right)

        nums = []
        dfs(root)
        return " ".join(map(str, nums))

    def deserialize(self, data: str) -> Optional[TreeNode]:
        """Decodes your encoded data to tree."""

        def dfs(mi: int, mx: int) -> Optional[TreeNode]:
            nonlocal i
            if i == len(nums) or not mi <= nums[i] <= mx:
                return None
            x = nums[i]
            root = TreeNode(x)
            i += 1
            root.left = dfs(mi, x)
            root.right = dfs(x, mx)
            return root

        nums = list(map(int, data.split()))
        i = 0
        return dfs(-inf, inf)

Java版：

public class Codec {
    private int i;
    private List<String> nums;
    private final int inf = 1 << 30;

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        nums = new ArrayList<>();
        dfs(root);
        return String.join(" ", nums);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || "".equals(data)) {
            return null;
        }
        i = 0;
        nums = Arrays.asList(data.split(" "));
        return dfs(-inf, inf);
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        nums.add(String.valueOf(root.val));
        dfs(root.left);
        dfs(root.right);
    }

    private TreeNode dfs(int mi, int mx) {
        if (i == nums.size()) {
            return null;
        }
        int x = Integer.parseInt(nums.get(i));
        if (x < mi || x > mx) {
            return null;
        }
        TreeNode root = new TreeNode(x);
        ++i;
        root.left = dfs(mi, x);
        root.right = dfs(x, mx);
        return root;
    }
}