LeetCode-Binary Tree Vertical Order Traversal

LiBlog發表於2016-08-01
LeetCode

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:

 

  1. Given binary tree [3,9,20,null,null,15,7],
       3
  /\
 /  \
 9  20
    /\
   /  \
  15   7

    return its vertical order traversal as:

    [
  [9],
  [3,15],
  [20],
  [7]
]
  2. Given binary tree [3,9,8,4,0,1,7],
         3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7

    return its vertical order traversal as:

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

 

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