Problem

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.

Solution

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;
        return dfs(root, 0, 1, new ArrayList<Integer>());
    }
    private int dfs(TreeNode root, int level, int id, List<Integer> leftnodes) {
        if (root == null) return 0;
        if (leftnodes.size() <= level) leftnodes.add(id);
        int left = dfs(root.left, level+1, id*2, leftnodes);
        int right = dfs(root.right, level+1, id*2+1, leftnodes);
        return Math.max(id-leftnodes.get(level)+1, Math.max(left, right));
    }
}

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