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.

题目理解：

给定一棵二叉树，如图，二叉树的每一个节点都有行坐标和列坐标，输出每一个列中的节点的值，按照行坐标递增顺序排列，如果有两个值的行列坐标都相同，那么按照值的大小排序

解题思路：

用字典表示二维数组，第一维表示列，第二维表示行，每一个元素都是一个链表，使用递归的方式遍历二叉树中的每一个节点，遍历的时候就附带上行列坐标，将每一个值都存储在对应的链表中，遍历完成之后按照要求组织答案即可

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeMap<Integer, TreeMap<Integer, List<Integer>>> map;
    public void helper(TreeNode root, int row, int col){
        if(root == null)
            return;
        if(!map.containsKey(col))
            map.put(col, new TreeMap<Integer, List<Integer>>());
        TreeMap<Integer, List<Integer>> it = map.get(col);
        if(!it.containsKey(row))
            it.put(row, new ArrayList<Integer>());
        it.get(row).add(root.val);
        helper(root.left, row + 1, col - 1);
        helper(root.right, row + 1, col + 1);
    }

    public List<List<Integer>> verticalTraversal(TreeNode root) {
        map = new TreeMap<>();
        helper(root, 0, 0);
        List<List<Integer>> res = new ArrayList<>();
        for(int col : map.keySet()){
            TreeMap<Integer, List<Integer>> it = map.get(col);
            List<Integer> list = new ArrayList<>();
            for(int row : it.keySet()){
                List<Integer> cur = it.get(row);
                Collections.sort(cur);
                list.addAll(cur);
            }
            res.add(list);
        }
        return res;
    }
}