662. Maximum Width of Binary Tree

problem description

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.

algorithm thought

这题是求每一层的宽度。如果要求宽度，关键是要得到最左和最右节点的位置，然后将两个位置相减就行。关键是如何安排每个节点的值。这里可以用类似heap的表示法。

如果first index base 1，它的子树就是 2index , 2index+1. 如果first index base 0，它的子树就是 2index+1 , 2index+2

然后用DFS或者是BFS遍历数，将最右的值减去最左的值即可。但是如果就这样，不加处理，会在最后两个case失败。因为我们用int类型表达index，这样树最多只能有32层。如果用long类型，树也只能有64层，这两种类型都不能通过测试。

于是我参考了下discuss，发现有人在进入下一层的时候，将index module INT_MAX。保证不溢出。修改代码之后发现能通过。

code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int widthOfBinaryTree(TreeNode* root) {
        if(root==NULL)
            return 0;
        vector<long> left;
        return help(root,left,1,0);
    }
    long help(TreeNode*root,vector<long>&left,long pos,int level){
        if(root==NULL)
            return 0;
        if(level>=left.size())
            left.push_back(pos);
        //cout<<pos<<'-'<<left[level]<<' ';
        long le=help(root->left,left,(2*pos)%INT_MAX,level+1);
        return max(pos-left[level]+1,max(le,help(root->right,left,(2*pos+1)%INT_MAX,level+1)));
    }
};

algorithm analysis

DFS遍历树，每次遍历中间时间复杂度为O(1)，最后时间复杂度O(n)