987. Vertical Order Traversal of a Binary Tree & 314

Problem Statement

link: LeetCode.cn LeetCode 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:

Input: root = [3,9,20,null,null,15,7] Output: [[9],[3,15],[20],[7]]

Input: root = [1,2,3,4,5,6,7] Output: ` [[4],[2],[1,5,6],[3],[7]]`

Solution Approach

Algorithm

Implement

    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: Optional[TreeNode]) -> List[List[int]]:
        # {level:(row, value)}  row = traversal_level 
        dic = defaultdict(list)
        def dfs(node,level,row):
            if node:
                dic[level].append((row,node.val))
                dfs(node.left, level - 1, row + 1)
                dfs(node.right, level + 1, row + 1)
        res = []
        dfs(root, 0, 0)
        for level, v_pair in sorted(dic.items()):
            res.append([value for row, value in sorted(v_pair)])
        return res
class Solution:
    def verticalTraversal(self, root: Optional[TreeNode]) -> List[List[int]]:
        # {level:(row, value)}  row = traversal_level 
        dic = defaultdict(list)
        q = [(root, 0, 0)]
        while q:
            node, level, row = q.pop()
            dic[level].append((row, node.val))
            if node.left:
                q.append((node.left, level - 1, row + 1))
            if node.right:
                q.append((node.right, level + 1, row + 1))
        res = []
        for level, r_v in sorted(dic.items()):
            values = [v for r, v in sorted(r_v)]
            res.append(values)
        return res

    
    

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