LeetCode-Binary Tree Vertical Order Traversal

Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column).

If two nodes are in the same row and column, the order should be from left to right.

Examples:

Given binary tree [3,9,20,null,null,15,7],

return its vertical order traversal as:

[
  [9],
  [3,15],
  [20],
  [7]
]

Given binary tree [3,9,8,4,0,1,7],

return its vertical order traversal as:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]

Given binary tree [3,9,8,4,0,1,7,null,null,null,2,5] (0's right child is 2 and 1's left child is 5),
```
     3
    /
   /  
   9   8
  /  /
 /  /  
 4  01   7
    /
   /  
   5   2
```
return its vertical order traversal as:
```
[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]
```

Analysis:

Note that in the second example, 2 is in the left subtree of root, but will after 8 in the results. So simple traverse may not work. We can use BFS with index recording.

Solution:

 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 public class Solution {
11     public List<List<Integer>> verticalOrder(TreeNode root) {
12         List<List<Integer>> resList = new ArrayList<List<Integer>>();
13         if (root==null) return resList;
14 
15         List<TreeNode> nodeList = new ArrayList<TreeNode>();
16         List<Integer> indexList = new ArrayList<Integer>();
17         int cur = 0, min = 0, max=0;
18         nodeList.add(root);
19         indexList.add(0);
20     
21         while (cur<nodeList.size()){
22             TreeNode curNode = nodeList.get(cur);
23             int curIndex = indexList.get(cur);
24             if (curIndex < min) min = curIndex;
25             if (curIndex > max) max = curIndex;
26 
27             if (curNode.left!=null){
28                 nodeList.add(curNode.left);
29                 indexList.add(curIndex-1);
30             }
31             if (curNode.right!=null){
32                 nodeList.add(curNode.right);
33                 indexList.add(curIndex+1);
34             }
35             cur++;
36         }
37         
38         for (int i=min;i<=max;i++) resList.add(new LinkedList<Integer>());
39         
40         for (int i=0;i<nodeList.size();i++){
41             resList.get(indexList.get(i)-min).add(nodeList.get(i).val);
42         }
43 
44         
45         return resList;
46     }
47         
48 }

相关阅读:
Android在代码中获取应用签名
Android之密码的显示与隐藏
如何把本地项目上传到Github
Android之移动热修复
Android之apk优化
Android之内存泄漏
Java类加载顺序
mysql系列-⼀条SQL查询语句是如何执⾏的？
java操作redis（jedis）常用方法示例
mybatis拦截器实现通用权限字段添加

原文地址：https://www.cnblogs.com/lishiblog/p/5725751.html