987. Vertical Order Traversal of a Binary Tree

Given a binary tree, return the vertical order traversal of its nodes values.

For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).

Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).

If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.

Return an list of non-empty reports in order of X coordinate. Every report will have a list of values of nodes.

Example 1:

Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation: 
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).

Example 2:

Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation: 
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6.

Note:

1. The tree will have between 1 and 1000 nodes.

2. Each node's value will be between 0 and 1000.

// Level Order BFS
vector<vector<int>> verticalTraversal(TreeNode* root) { // time: O(nlogn); space: O(n)
    vector<vector<int> > res;
    if (!root) return res;
    map<int, vector<pair<int, int> > > mp; // x->vector<pair<y, value> >
    int x = 0, y = 0;
    queue<pair<TreeNode*, pair<int, int> > > q; // pair<TreeNode*, pair<x, y> >
    q.push({root, {x, y} });
    while (!q.empty()) {
        auto t = q.front(); q.pop();
        TreeNode* node = t.first;
        const pair<int, int>& p = t.second;
        mp[p.first].push_back({p.second, node->val});
        if (node->left) q.push({node->left, {p.first - 1, p.second + 1} });
        if (node->right) q.push({node->right, {p.first + 1, p.second + 1} });
    }
    int n = mp.size();
    res.resize(n);
    int idx = 0;
    for (auto& e : mp) {
        auto& tmp_vec = e.second;
        sort(tmp_vec.begin(), tmp_vec.end());
        for (const pair<int, int>& p : tmp_vec) {
            res[idx].push_back(p.second);
        }
        ++idx;
    }
    return res;
}

// Recursion
void helper(TreeNode* node, int x, int y, unordered_map<int, unordered_map<int, set<int> > >& mp) {
    if (!node) return;
    mp[x][y].insert(node->val);
    helper(node->left, x - 1, y + 1, mp);
    helper(node->right, x + 1, y + 1, mp);
}
vector<vector<int>> verticalTraversal(TreeNode* root) { // time: O(nlogn); space: O(n)
    vector<vector<int> > res;
    unordered_map<int, unordered_map<int, set<int> > > mp;
    helper(root, 0, 0, mp);
    for (int x = -999; x <= 999; ++x) {
        if (!mp.count(x)) continue;
        res.push_back(vector<int>() );
        for (int y = 0; y < 1000; ++y) {
            if (!mp[x].count(y)) continue;
            res.back().insert(res.back().end(), mp[x][y].begin(), mp[x][y].end());
        }
    }
    return res;
}

// Recursion
void helper(TreeNode* node, int x, int y, map<pair<int, int>, set<int> >& mp, int& min_x, int& max_x) {
    if (!node) return;
    min_x = min(min_x, x);
    max_x = max(max_x, x);
    mp[{y, x}].insert(node->val);
    helper(node->left, x - 1, y + 1, mp, min_x, max_x);
    helper(node->right, x + 1, y + 1, mp, min_x, max_x);
}
vector<vector<int>> verticalTraversal(TreeNode* root) { // time: O(nlogn); space: O(n)
    vector<vector<int> > res;
    if (!root) return res;
    int min_x = numeric_limits<int>::max(), max_x = numeric_limits<int>::min();
    map<pair<int, int>, set<int> > mp; // {y, x} -> {vals}
    helper(root, 0, 0, mp, min_x, max_x);
    res.resize(max_x - min_x + 1);
    for (const auto& e : mp) {
        int x = e.first.second - min_x;
        res[x].insert(end(res[x]), begin(e.second), end(e.second));
    }
    return res;
}