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.
Input: [3,9,20,null,null,15,7] Output: [,[3,15],,] 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).
Input: [1,2,3,4,5,6,7] Output: [,,[1,5,6],,] 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.
- The tree will have between 1 and 1000 nodes.
- Each node's value will be between 0 and 1000.
计算每个节点的 (x, y) 坐标，然后按 x 轴的值排序， 遍历排序后的结果，x 轴值相等的节点在同一条垂直线上可以归为一组。
这个方法的 Python 代码类似下面这样:
# 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): self._positions =  self._collect_positions(root, 0, 0) vals =  last_x = -1001 for x, y, val in sorted(self._positions): # x 轴值相同的归为一组 if x != last_x: vals.append() last_x = x vals[-1].append(val) return vals def _collect_positions(self, root, x, y): if root is None: return self._positions.append((x, y, root.val)) # 之所以是 y + 1 而不是 y - 1 是为了实现排序后低层级的在前面 self._collect_positions(root.left, x - 1, y + 1) self._collect_positions(root.right, x + 1, y + 1)