<reponame>sugia/leetcode
'''
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:

The tree will have between 1 and 1000 nodes.
Each node's value will be between 0 and 1000.
'''

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def verticalTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        # col: node.val
        dic = {}
        
        vec = [(root, 0)]
        while vec:
            next_vec = []
            for node, col in vec:
                if col not in dic:
                    dic[col] = [node.val]
                else:
                    dic[col].append(node.val)
                if node.left:
                    next_vec.append((node.left, col-1))
                if node.right:
                    next_vec.append((node.right, col+1))
            vec = sorted(next_vec, key = lambda x: (x[1], x[0].val))
            
        res = []
        for col in sorted(dic.keys()):
            res.append(dic[col])
        return res

'''
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:

The tree will have between 1 and 1000 nodes.
Each node's value will be between 0 and 1000.
'''

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def verticalTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        # col: node.val
        dic = {}
        
        vec = [(root, 0)]
        while vec:
            next_vec = []
            for node, col in vec:
                if col not in dic:
                    dic[col] = [node.val]
                else:
                    dic[col].append(node.val)
                if node.left:
                    next_vec.append((node.left, col-1))
                if node.right:
                    next_vec.append((node.right, col+1))
            vec = sorted(next_vec, key = lambda x: (x[1], x[0].val))
            
        res = []
        for col in sorted(dic.keys()):
            res.append(dic[col])
        return res
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:  The tree will have between 1 and 1000 nodes. Each node's value will be between 0 and 1000. # Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None :type root: TreeNode         :rtype: List[List[int]] # col: node.val