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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.
 

Related Topics:
Hash Table, Tree

Solution 1.

Use a map<int, map<int, multiset<int>>> m to store the values -- m[node->x][node->y].insert(node->val). (Using set instead of multiset can also pass this problem. I guess LeetCode uses the node values as IDs and assumes the uniqueness of the values. I used multiset here to be safe.)

In this way, the values are sorted first in asending order of the x values, then in asending order of y values, then in asending order of node values.

Note that we shouldn't sort the values with the same X values all together, we should only sort them if they have the same position, i.e. when both their x and y values are equal.

// OJ: https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
    map<int, map<int, multiset<int>>> m;
    void dfs(TreeNode *root, int x, int y) {
        if (!root) return;
        m[x][y].insert(root->val);
        dfs(root->left, x - 1, y + 1);
        dfs(root->right, x + 1, y + 1);
    }
public:
    vector<vector<int>> verticalTraversal(TreeNode* root) {
        dfs(root, 0, 0);
        vector<vector<int>> ans;
        for (auto &[x, mm] : m) {
            ans.emplace_back();
            for (auto &[y, vals] : mm) {
                for (int n : vals) ans.back().push_back(n);
            }
        }
        return ans;
    }
};