LeetCode 987. Vertical Order Traversal of a Binary Tree

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题目

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:

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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:

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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:

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

思路

DFS.
I have a dictionary with the keys: x and the value: a tuple of (y, root.val). I then sort the dictionary based on keys and sort it again based on y.

代码

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def verticalTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        dic = collections.defaultdict(list)
        self.dfs(0, 0, root, dic)
        res = []
        for i in sorted(dic.keys()):
            tmp = []
            for j in sorted(dic[i]):
                tmp.append(j[1])
            res.append(tmp)
        return res
        
    
    def dfs(self, x, y, root, dic):
        if not root:
            return 
        dic[x].append((y, root.val))
        self.dfs(x-1, y+1, root.left, dic)
        self.dfs(x+1, y+1, root.right, dic)