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.
Runtime: 8 ms, faster than 28.24% of C++ online submissions for Maximum Width of Binary Tree.
看了一下更快的解法用的是递归，但是思路和我的是一样的。就是给每一个点都标上序号。
最快的解法竟然是用while嵌套BFS，看来递归花掉了很多时间，但我这个可能是由于有拷贝（nextq,q）。
/**
* 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) {
queue<pair<TreeNode*, int>> q;
q.push({root, 1});
int ret = 0;
while(!q.empty()){
int minval = INT_MAX, maxval = INT_MIN;
queue<pair<TreeNode*,int>> nextq;
while(!q.empty()){
auto p = q.front(); q.pop();
minval = min(minval, p.second); maxval = max(maxval, p.second);
if(p.first->left) nextq.push({p.first->left, 2*p.second});
if(p.first->right) nextq.push({p.first->right, 2*p.second+1});
}
ret = max(ret, maxval - minval);
q = nextq;
}
return ret+1;
}
};