Leetcode-Challenges-Python/August_LeetCode_Challenge/Vertical_Order_Traversal_Of_Binary_Tree.py at master · GSNCodes/Leetcode-Challenges-Python · GitHub

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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:

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:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:


    def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
        nodes = defaultdict(list)
        def dfs(root, x, y):
            if not root:
                return
            nodes[x].append((root.val, y))
            dfs(root.left, x-1, y+1)
            dfs(root.right, x+1, y+1)

        dfs(root, 0, 0)
        print(nodes)
        return [[n[0] for n in sorted(nodes[k], key=lambda x:(x[1], x[0]))]
                for k in sorted(nodes.keys())]


class Solution:
    def vertical_transversal_helper(root):
                queue = deque()
                queue.append((root, 0, 0))

                while queue:
                    node, row, col  = queue.popleft()
                    if node is not None:
                        nodes.append((col, row, node.val))
                        queue.append((node.left, row + 1, col - 1))
                        queue.append((node.right, row + 1, col + 1))

            nodes = []
            vertical_transversal_helper(root)
            nodes.sort()
            ret = OrderedDict()
            for col, row, value in nodes:
                if col in ret:
                    ret[col].append(value)
                else:
                    ret[col] = [value]
            return ret.values()