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
and1000
.
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;但是这个题的题设写的非常不清楚,根据test case,实际的要求是
- 如果是正常情况,自然就是从左往右按column输出所有节点;在每个column中,节点按照自身到根节点的距离由近到远输出
- 如果有两个节点的偏移量相同(应该是在同一个column里或一个是左孩子一个是右孩子),而且他们之于根节点的距离也相同,那么再按照node.val从小到大排序,比如例子中的5就在6的前面
这里我提供一个BFS的做法,思路跟314题也很像。这里我还是需要用到两个queue,一个存遍历的所有节点,一个存每个节点相对于根节点root的偏移量;我还需要一个总的hashmap和一个临时的hashmap temp。
之后按BFS的思路遍历整棵树,首先我们需要记录每一层的节点个数size,同时我们需要对同一层的节点进行一个局部的统计,用一个临时的hashmap temp记录每一层的节点的偏移量和节点信息。当前层遍历完毕之后,对于这个temp,我们把所有的节点拿出来并排序,把排序好的list再加入总的hashmap里,注意拿的方式是很巧妙的,因为我们在遍历一个hashmap的keySet的时候,key是按照被加入的顺序被读出来的,而且我们在存的时候,因为是BFS遍历所以基本顺序是根节点,左孩子,右孩子。所以key从temp里被读出来的基本顺序也是根节点,左孩子,右孩子,左孩子的孩子,右孩子的孩子。这个顺序也就隐性地满足了偏移量相同的时候的排序要求。其他环节跟314题基本相同。
时间O(nlogn)
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() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public List<List<Integer>> verticalTraversal(TreeNode root) { 18 List<List<Integer>> res = new ArrayList<>(); 19 // corner case 20 if (root == null) { 21 return res; 22 } 23 24 // normal case 25 int min = 0; 26 int max = 0; 27 HashMap<Integer, List<Integer>> map = new HashMap<>(); 28 Queue<TreeNode> queue = new LinkedList<>(); 29 Queue<Integer> cols = new LinkedList<>(); 30 queue.offer(root); 31 cols.offer(0); 32 while (!queue.isEmpty()) { 33 int size = queue.size(); 34 HashMap<Integer, List<Integer>> temp = new HashMap<>(); 35 for (int i = 0; i < size; i++) { 36 TreeNode cur = queue.poll(); 37 int index = cols.poll(); 38 if (!temp.containsKey(index)) { 39 temp.put(index, new ArrayList<>()); 40 } 41 temp.get(index).add(cur.val); 42 min = Math.min(min, index); 43 max = Math.max(max, index); 44 if (cur.left != null) { 45 queue.offer(cur.left); 46 cols.offer(index - 1); 47 } 48 if (cur.right != null) { 49 queue.offer(cur.right); 50 cols.offer(index + 1); 51 } 52 } 53 // 在这里sort是因为深度不同的时候,深度小的在前 54 // 深度相同的时候val小的在前 55 for (int key : temp.keySet()) { 56 if (!map.containsKey(key)) { 57 map.put(key, new ArrayList<>()); 58 } 59 List<Integer> list = temp.get(key); 60 Collections.sort(list); 61 map.get(key).addAll(list); 62 } 63 } 64 65 for (int i = min; i <= max; i++) { 66 List<Integer> list = map.get(i); 67 res.add(list); 68 } 69 return res; 70 } 71 }
相关题目
314. Binary Tree Vertical Order Traversal