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: alt

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: alt

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.

Solution: Ordered Map+ Ordered Set¶

Time complexity: O(nlogn) Space complexity: O(n)

class Solution {
public:
  vector<vector<int>> verticalTraversal(TreeNode* root) {
    if (!root) return {};
    int min_x = INT_MAX;
    int max_x = INT_MIN;
    map<pair<int, int>, set<int>> h; // {y, x} -> {vals}
    traverse(root, 0, 0, h, min_x, max_x);
    vector<vector<int>> ans(max_x - min_x + 1);
    for (const auto& m : h) {
      int x = m.first.second - min_x;
      ans[x].insert(end(ans[x]), begin(m.second), end(m.second));
    }
    return ans;
  }
private:
  void traverse(TreeNode* root, int x, int y,
                map<pair<int, int>, set<int>>& h,
                int& min_x,
                int& max_x) {
    if (!root) return;
    min_x = min(min_x, x);
    max_x = max(max_x, x);
    h[{y, x}].insert(root->val);
    traverse(root->left, x - 1, y + 1, h, min_x, max_x);
    traverse(root->right, x + 1, y + 1, h, min_x, max_x);
  }
};