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.

solution

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> verticalTraversal(TreeNode* root) {
        x_min_ = INT_MAX;
        x_max_ = INT_MIN;
        
        inorderTraversal(root, 0, 0);
        
        vector<vector<int>> res(x_max_-x_min_+1);
        for (auto it: map_) {it.first.second);
            int x = it.first.second - x_min_;
            res[x].insert(end(res[x]), begin(it.second), end(it.second));
        }
        return res;
    }
private:
    map<pair<int, int>, set<int>> map_;
    int x_min_, x_max_;
    void inorderTraversal(TreeNode* node, int x, int y) {
        if (!node) return;
        x_min_ = min(x_min_, x);
        x_max_ = max(x_max_, x);
        map_[{y, x}].insert(node->val);
        inorderTraversal(node->left, x-1, y+1);
        inorderTraversal(node->right, x+1, y+1);
    }
};

用pair (y, x) map to int set。
在traverse時的y與題目相反，用增加的，可以讓map以y優先小到大放置(即樹的上到下)，將traverse結果一一存入map中，同時記錄x值的最小、最大值，以便求範圍。

再根據x值的範圍，建出res array，再把map中的元素一一放入res中。