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  • [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;
    }
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  • 原文地址:https://www.cnblogs.com/hellowooorld/p/7399335.html
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