https://leetcode-cn.com/problems/vertical-order-traversal-of-a-binary-tree/

problems

Given the root of a binary tree, calculate the vertical order traversal of the binary tree.

For each node at position (row, col), its left and right children will be at positions (row + 1, col - 1) and (row + 1, col + 1) respectively. The root of the tree is at (0, 0).

The vertical order traversal of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values.

Return the vertical order traversal of the binary tree.

example

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation:
Column -1: Only node 9 is in this column.
Column 0: Nodes 3 and 15 are in this column in that order from top to bottom.
Column 1: Only node 20 is in this column.
Column 2: Only node 7 is in this column.

Example 2:

Input: root = [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation:
Column -2: Only node 4 is in this column.
Column -1: Only node 2 is in this column.
Column 0: Nodes 1, 5, and 6 are in this column.
          1 is at the top, so it comes first.
          5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6.
Column 1: Only node 3 is in this column.
Column 2: Only node 7 is in this column.

Example 3:

Input: root = [1,2,3,4,6,5,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation:
This case is the exact same as example 2, but with nodes 5 and 6 swapped.
Note that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 1000

solution

思路：

1. 把所有节点的坐标遍历记录下来。
2. 按照纵坐标顺序，行坐标从小到大
3. 细节，同样的坐标，值从小到到。 可以用嵌套的TreeMap 纵坐标为外层 key，横坐标为内层坐标key，这样遍历坐标符合从小到大。由于坐标可能重叠，用List 保存值，读取的时候需要注意顺序。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        TreeMap<Integer, TreeMap<Integer, List<Integer>>> map = new TreeMap<>();
        dfs(root,0, 0, map);
        //
        List<List<Integer>> ret = new ArrayList<>();
        for (Map.Entry<Integer, TreeMap<Integer, List<Integer>>> entry: map.entrySet()) {
            List<Integer> tmp = new ArrayList<>();
            for (Map.Entry<Integer, List<Integer>> entry1: entry.getValue().entrySet()) {
                List<Integer> list = entry1.getValue();
                Collections.sort(list);
                tmp.addAll(list);
            }
            if (tmp.size() >0) {
                ret.add(tmp);
            }
        }
        return ret;
    }
    public void dfs(TreeNode n, int x, int y, TreeMap<Integer, TreeMap<Integer, List<Integer>>> m) {
        if (n == null) {
            return;
        }
        //vist n;
        m.putIfAbsent(y, new TreeMap<>());
        TreeMap<Integer, List<Integer>> sub = m.get(y);
        sub.putIfAbsent(x, new ArrayList<Integer>());
        m.get(y).get(x).add(n.val);
        // vist left;
        dfs(n.left, x+1, y-1, m);
        // vist right;
        dfs(n.right, x+1, y+1, m);
    }
}