Description
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.
分析
题目的意思是:求二叉树的竖直遍历的输出,同一竖直的数还要按照从小到大进行排列。这道题比较新,我也没做出来,参考了一下别人的做法,还是递归遍历,在存入数的时候要记录的层数,和垂直的索引。这样用一个字典存储起来,键是竖直的索引,值为当前节点的值和所在的深度。这样的话在最后排序一下就行了。思路挺新的,我做不出来,摊牌了哈哈哈。
代码
# 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 solve(self,col,depth,root):
if(root is None):
return
self.res[col].append([depth,root.val])
self.solve(col-1,depth+1,root.left)
self.solve(col+1,depth+1,root.right)
def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
self.res = defaultdict(list)
self.solve(0,0,root)
res2=[]
for col in sorted(self.res.keys()):
sorted_res=sorted(self.res[col],key=lambda x:(x[0],x[1]))
res2.append([item[1] for item in sorted_res])
return res2
参考文献
[LeetCode] 17 lines short Python code with explanation. Beats 98%