1028

1028. Recover a Tree From Preorder Traversal (Hard)

We run a preorder depth first search on the root of a binary tree.

At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node.  (If the depth of a node is D, the depth of its immediate child is D+1.  The depth of the root node is 0.)

If a node has only one child, that child is guaranteed to be the left child.

Given the output S of this traversal, recover the tree and return its root.

 

Example 1:

Input: "1-2--3--4-5--6--7"
Output: [1,2,5,3,4,6,7]

Example 2:

Input: "1-2--3---4-5--6---7"
Output: [1,2,5,3,null,6,null,4,null,7]

 

Example 3:

Input: "1-401--349---90--88"
Output: [1,401,null,349,88,90]

 

Note:

• The number of nodes in the original tree is between 1 and 1000.
• Each node will have a value between 1 and 10^9.

Solution 1. Stack

// OJ: https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(S)
// Space: O(H) where H is depth of tree
class Solution {
private:
    TreeNode *readNode(string &S, int &i) {
        int n = 0;
        for (; i < S.size() && isdigit(S[i]); ++i) n = 10 * n + S[i] - '0';
        return new TreeNode(n);
    }
    int readDash(string &S, int &i) {
        int cnt = 0;
        for (; i < S.size() && S[i] == '-'; ++i, ++cnt);
        return cnt;
    }
public:
    TreeNode* recoverFromPreorder(string S) {
        int i = 0, N = S.size();
        auto root = readNode(S, i);
        stack<TreeNode*> s;
        s.push(root);
        while (i < S.size()) {
            int dep = readDash(S, i);
            auto node = readNode(S, i);
            while (dep < s.size()) s.pop();
            auto p = s.top();
            if (p->left) p->right = node;
            else p->left = node;
            s.push(node);
        }
        return root;
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(S)
// Space: O(H)
class Solution {
    int curLevel;
    TreeNode *curNode;
    void dfs(istringstream &in, TreeNode *parent, int level) {
        if (in.eof()) return;
        curLevel = 0;
        for (; in.peek() == '-'; in.get(), ++curLevel);
        int n;
        in >> n;
        curNode = new TreeNode(n);
        while (curNode && curLevel == level) {
            if (!parent->left) parent->left = curNode;
            else parent->right = curNode;
            auto node = curNode;
            curNode = NULL;
            dfs(in, node, level + 1);
        }
    }
public:
    TreeNode* recoverFromPreorder(string S) {
        istringstream in(S);
        int n;
        in >> n;
        auto root = new TreeNode(n);
        dfs(in, root, 1);
        return root;
    }
};