LeetCode 987. Vertical order transverse of a binary tree - Binary Tree Series Question 7

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 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

This question is related to 314. Binary Tree Vertical Order Traversal It's almost exactly the same , The change is that nodes with the same row number and column number need to be arranged from small to large according to the node value . Problem solving ideas are also in 314. Binary Tree Vertical Order Traversal Slightly changed on the basis of , When doing horizontal sequence traversal , Take one more line number , That is, the line number and node value of each node are represented by tuples tuple(row, val) Form into the column array . In this way, at the end, just sort each column , Because it is traversed in horizontal order, the line number is from small to large , Therefore, sorting will only change the order of node values with the same line number , Just meet the requirements .

class Solution:
    def verticalTraversal(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        
        llist, rlist = [], []
        
        lindex, rindex = 0, 1
        
        q = deque([(root, 0, 0)])
        while q:
            n = len(q)
            for i in range(n):
                node, row, col = q.popleft()
                if col <= 0:
                    if len(llist) > -col:
                        llist[-col].append((row, node.val))
                    else:
                        llist.append([(row, node.val)])
                else:
                    if len(rlist) > col - 1:
                        rlist[col - 1].append((row, node.val))
                    else:
                        rlist.append([(row, node.val)])                                  
                if node.left:
                    q.append((node.left, row + 1, col - 1))
                if node.right:
                    q.append((node.right, row + 1, col + 1))
                    
        llist.reverse()
        wlist = llist + rlist
        
        res = []
        for arr in wlist:
            arr.sort()
            res.append([v[1] for v in arr])
            
        return res