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LeetCode 987 Vertical Order Traversal Of A Binary Tree - Medium

987. Vertical Order Traversal of a Binary Tree -- Medium

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987. Vertical Order Traversal Of A Binary Tree — Medium

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Problem

  1. Vertical Order Traversal of a Binary Tree -- Medium

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.

Solution

### BFS solution:
import collections
# 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]]:
        map = self.bfs(root)
        res = []
        for key in sorted(map.keys()):
            res.append( [x[1] for x in sorted(map[key])])
        return res
        
    def bfs(self, root):
        queue = collections.deque([(root, 0, 0)])
        map = collections.defaultdict(list)
        while queue:
            node, x, y = queue.popleft()
            map[x].append((-y, node.val))
            if node.left:
                queue.append((node.left, x - 1, y - 1))
            if node.right:
                queue.append((node.right, x + 1, y - 1))
        return map
 
### DFS solution:
import collections
# 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]]:
        self.map = collections.defaultdict(list)
        self.dfs(root, 0, 0)
        res = []
        for key in sorted(self.map.keys()):
            res.append([x[1] for x in sorted(self.map[key])])
        return res
        
    def dfs(self, root, x, y):
        if not root:
            return
        self.map[x].append((-y, root.val))
        self.dfs(root.left, x - 1, y - 1)
        self.dfs(root.right, x + 1, y - 1)

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