[leetcode-662-Maximum Width of Binary Tree]

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).

思路：

层次遍历，然后每一个结点对应一个标号，每一层的宽度用最右边的标号-最左边标号即可。

int widthOfBinaryTree(TreeNode* root)
{
    if( root == NULL ) return 0;
    queue< TreeNode*  > qu;
    map< TreeNode*, int > mp;

    qu.push( root );
    int maxW = 0xc0c0c0c0;

    int numL = -1, numR = -1;

    while( !qu.empty() )
    {
        int n = qu.size();

        for( int i = 0; i < n; i++ )
        {
            TreeNode* tmp = qu.front();
            qu.pop();
            if( i == 0 )
            {
                numL = mp[tmp];
            }
            if( i == n-1 )
            {
                numR = mp[tmp];
            }

            if( tmp->left != NULL )
            {
                qu.push( tmp->left );
                mp[tmp->left] = mp[tmp] * 2;
            }
            if( tmp->right != NULL )
            {
                qu.push( tmp->right );
                mp[tmp->right] = mp[tmp] * 2 + 1;
            }
        }
        maxW = max( maxW, numR - numL + 1 );
    }
    return maxW;
}