662. Maximum Width of Binary Tree.md

662. Maximum Width of Binary Tree

Difficulty : Medium

Related Topics : Tree

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leetcode Daily Challenge on July 9th, 2020.

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Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

Example 1:

Input: 

           1
         /   \
        3     2
       / \     \  
      5   3     9 

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Example 2:

Input: 

          1
         /  
        3    
       / \       
      5   3     

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).

Example 3:

Input: 

          1
         / \
        3   2 
       /        
      5      

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).

Example 4:

Input: 

          1
         / \
        3   2
       /     \  
      5       9 
     /         \
    6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

Note

• Answer will in the range of 32-bit signed integer.

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Solution

• mine
  • Java
    • DFS Runtime: 2 ms, faster than 45.85%, Memory Usage: 40.2 MB, less than 13.33% of Java online submissions

      // O(N)time 
      // O(N)space
      int res = 0;
      public int widthOfBinaryTree(TreeNode root) {
          Map<Integer, int[]> map = new HashMap<>(); 
          helper(root, 1, 1, map);
          return res;
      }
      
      void helper(TreeNode node, int depth, int num, Map<Integer, int[]> map){
          if(node == null){
              return;
          }
          int[] record = map.getOrDefault(depth, new int[]{0,0});
          if(record[0] == 0){
              record[0] = num;
          }else{
              int t = num;
              if(num < record[0]){
                  t = record[0];
                  record[0] = num;
              }
              if(record[1] == 0 || t > record[1]){
                  record[1] = t;
              }
          }
          res = Math.max(res, record[1] == 0 ? 1 : record[1] - record[0] + 1);
          map.put(depth, record);
          if(node.left != null){
              helper(node.left, depth + 1, num * 2, map);
          }
          if(node.right != null){
              helper(node.right, depth + 1, num * 2 + 1, map);
          }
      }
      

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• the most votes
  • DFS Runtime: 2 ms, faster than 45.85%, Memory Usage: 40 MB, less than 17.64% of Java online submissions

    // O(N)time
    // O(N)space
    public int widthOfBinaryTree(TreeNode root) {
        return dfs(root, 0, 1, new ArrayList<Integer>(), new ArrayList<Integer>());
    }
    
    public int dfs(TreeNode root, int level, int order, List<Integer> start, List<Integer> end){
        if(root == null)return 0;
        if(start.size() == level){
            start.add(order); end.add(order);
        }
        else end.set(level, order);
        int cur = end.get(level) - start.get(level) + 1;
        int left = dfs(root.left, level + 1, 2*order, start, end);
        int right = dfs(root.right, level + 1, 2*order + 1, start, end);
        return Math.max(cur, Math.max(left, right));
    }
    

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