Binary Tree Vertical Order Traversal 314

Description

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],

  3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   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),

return its vertical order traversal as: [ [4], [9,5], [3,0,1], [8,2], [7] ]

Hint

looking from top to bottom just like putting each node into right columns;

Method

hashmap to store each using queue to do BFS from root to both sides col - 1 col + 1 since map is used to store the relationship between col and sublist we can use traversal to calculate min, max to get all list index and then to BFS to avoid using map

Time & Space

O(n)

Code

public List<List<Integer>> verticalOrder(TreeNode root) {
           List<List<Integer>> ans = new ArrayList<List<Integer>>();
           if (root == null){
               return ans;
           }

           HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
           Queue<TreeNode> q = new LinkedList<TreeNode>();
           Queue<Integer> cols = new LinkedList<Integer>();


           q.offer(root);
           cols.offer(0);

           int min = 0;
           int max = 0;

           while (!q.isEmpty()){
                  TreeNode node = q.poll();
                  int col = cols.poll();
                  if (!map.containsKey(col)){
                      map.put(col, new ArrayList<Integer>());
                  }
                  map.get(col).add(node.val);
                  if (node.left != null){
                      q.offer(node.left);
                      cols.offer(col - 1);
                      min = Math.min(col - 1, min);
                  }
                  if (node.right != null){
                      q.offer(node.right);
                      cols.offer(col + 1);
                      max = Math.max(col + 1, max);
                  }
           }

           for (int i = min; i <= max; i++){
                ans.add(map.get(i));
           }
           return ans;
    }

int min = 0;
int max = 0;
public List<List<Integer>> verticalOrder(TreeNode root) {
       List<List<Integer>> ans = new ArrayList<List<Integer>>();
       if (root == null){
          return ans;
       }
       helper(root, 0);

       for (int i = min; i <= max; i++){
           ans.add(new ArrayList<Integer>());
       }

       Queue<TreeNode> q = new LinkedList<TreeNode>();
       Queue<Integer> cols = new LinkedList<Integer>();

       q.offer(root);
       cols.offer(-min);

       while (!q.isEmpty()){
              TreeNode node = q.poll();
              int col = cols.poll();

              ans.get(col).add(node.val);
              if (node.left != null){
                  q.offer(node.left);
                  cols.offer(col - 1);
              }
              if (node.right != null){
                  q.offer(node.right);
                  cols.offer(col + 1);
              }
      }
      return ans;
}

public void helper(TreeNode root, int col){
       if (root == null){
           return;
       }
       min = Math.min(col, min);
       max = Math.max(col, max);
       helper(root.left, col - 1);
       helper(root.right, col + 1);
}

2.5 Binary Tree Vertical Order Traversal 314