[Swift Weekly Contest 122]LeetCode987. 二叉树的垂序遍历 | Ve

发布时间：2020-12-14 05:04:31 所属栏目：百科来源：网络整理

导读：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 = -i

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,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],[1,6],[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 "[1,6]",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.

给定二叉树，按垂序遍历返回其结点值。

对位于?(X,Y)?的每个结点而言，其左右子结点分别位于?(X-1,Y-1)?和?(X+1,Y-1)。

把一条垂线从?X = -infinity?移动到?X = +infinity?，每当该垂线与结点接触时，我们按从上到下的顺序报告结点的值（?Y?坐标递减）。

如果两个结点位置相同，则首先报告的结点值较小。

按?X?坐标顺序返回非空报告的列表。每个报告都有一个结点值列表。

示例 1：

输入：[3,7]
输出：[[9],[7]]
解释： 
在不丧失其普遍性的情况下，我们可以假设根结点位于 (0,0)：
然后，值为 9 的结点出现在 (-1,-1)；
值为 3 和 15 的两个结点分别出现在 (0,0) 和 (0,-2)；
值为 20 的结点出现在 (1,-1)；
值为 7 的结点出现在 (2,-2)。

示例 2：

输入：[1,7]
输出：[[4],[7]]
解释：
根据给定的方案，值为 5 和 6 的两个结点出现在同一位置。
然而，在报告 "[1,6]" 中，结点值 5 排在前面，因为 5 小于 6。

提示：

树的结点数介于?1?和?1000?之间。
每个结点值介于?0?和?1000?之间。

Runtime:?20 ms

Memory Usage:?4.1 MB

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     var hi:[[Int]] = [[Int]]()
16     func verticalTraversal(_ root: TreeNode?) -> [[Int]] {
17         dfs(root,0,0)
18         hi.sort(by:sortArray)
19         var ret:[[Int]] = [[Int]]()
20         var i:Int = 0
21         while(i < hi.count)
22         {
23             var j:Int = i
24             while(j < hi.count && hi[j][1] == hi[i][1])
25             {
26                 j += 1
27             }
28             var item:[Int] = [Int]()
29             for k in i..<j
30             {
31                 item.append(hi[k][0])
32             }
33             ret.append(item)
34             i = j
35         }
36         return ret
37     }
38     
39     func sortArray(_ a:[Int],_ b:[Int]) -> Bool
40     {
41         if a[1] != b[1] {return a[1] < b[1]}
42         if a[2] != b[2] {return a[2] > b[2]}
43         return a[0] < b[0]
44     }
45     
46     func dfs(_ cur: TreeNode?,_ x:Int,_ y:Int)
47     {
48         if cur == nil {return}
49         hi.append([cur!.val,x,y])
50         dfs(cur!.left,x-1,y-1)
51         dfs(cur!.right,x+1,y-1)
52     }
53 }

（编辑：李大同）

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