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[LeetCode] 987. Vertical Order Traversal of a Binary Tree

时间:2020-05-13 12:52:18      阅读:59      评论:0      收藏:0      [点我收藏+]

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.

二叉树的垂序遍历。题意跟314题非常像,但是314只要求我们找到横坐标一样的元素,把他们合成一组;但是这个题是要求不光是横坐标相同的元素分在一组,同时要求要对坐标节点值排序,所以会用到treemap和PriorityQueue。treemap可以输出key是有序的hashmap,所以用来存储坐标;PriorityQueue可以存节点值。整体思路是用DFS前序遍历,遍历树不难,难的是要处理好数据怎么存。

时间O(nlogn) - 因为有PriorityQueue的排序

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<List<Integer>> verticalTraversal(TreeNode root) {
12         TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
13         dfs(root, 0, 0, map);
14         List<List<Integer>> list = new ArrayList<>();
15         for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) {
16             list.add(new ArrayList<>());
17             for (PriorityQueue<Integer> nodes : ys.values()) {
18                 while (!nodes.isEmpty()) {
19                     list.get(list.size() - 1).add(nodes.poll());
20                 }
21             }
22         }
23         return list;
24     }
25 
26     private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) {
27         if (root == null) {
28             return;
29         }
30         if (!map.containsKey(x)) {
31             map.put(x, new TreeMap<>());
32         }
33         if (!map.get(x).containsKey(y)) {
34             map.get(x).put(y, new PriorityQueue<>());
35         }
36         map.get(x).get(y).offer(root.val);
37         dfs(root.left, x - 1, y + 1, map);
38         dfs(root.right, x + 1, y + 1, map);
39     }
40 }

 

LeetCode 题目总结

[LeetCode] 987. Vertical Order Traversal of a Binary Tree

原文:https://www.cnblogs.com/cnoodle/p/12881028.html

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