Leetcode 987. Vertical Order Traversal of a Binary Tree
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:
Output: [,,,]
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:
Output: [,,,,]
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 "", 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.
# 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]:
res = []
if root == None:
return res
hashmap = collections.defaultdict(list)
queue = []
cols = []
max_val = 0
min_val = 0
level = 0
queue.append(root)
cols.append(0)
while queue:
size = len(queue)
for i in range(size):
curt = queue.pop(0)
col = cols.pop(0)
hashmap.append((level, curt.val))
if curt.left != None:
min_val = min(min_val, col - 1)
queue.append(curt.left)
cols.append(col - 1)
if curt.right != None:
max_val = max(max_val, col + 1)
queue.append(curt.right)
cols.append(col + 1)
level += 1
for i in range(min_val, max_val + 1):
res.append()])
return res
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