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.

Explain

這道題的意思是：給一個(gè)有著滿二叉樹結(jié)構(gòu)照激，但是有部分節(jié)點(diǎn)是null的樹，然后求出這個(gè)樹最寬的那一層的寬度。這個(gè)寬度就是每一層的最右和最左的距離的最大值。看起來不是很難做，有一個(gè)很暴力的方法就是把一層的節(jié)點(diǎn)都保存下來不管空不空，然后一個(gè)一個(gè)去遍歷找到最左和最右的下標(biāo)批销。但是我們作為一個(gè)coder洒闸，這么蠢的方法還是少用。這里要用到樹的一個(gè)特點(diǎn)均芽。每個(gè)節(jié)點(diǎn)的下標(biāo)如果是i丘逸，那么它的左節(jié)點(diǎn)是 2 * i, 右節(jié)點(diǎn)是 2 * i + 1。然后掀宋，我們只要將每個(gè)節(jié)點(diǎn)和它的位置保存下來深纲，做層序遍歷。開始一層的遍歷時(shí)劲妙，定義一個(gè)最左和一個(gè)最右的變量湃鹊，將這層第一個(gè)節(jié)點(diǎn)的位置保存下來，然后每遍歷這層的一個(gè)節(jié)點(diǎn)镣奋，將該節(jié)點(diǎn)的位置更新為最右變量的值币呵。這一層遍歷完后就可以得到這層的寬度，然后跟當(dāng)前最大的寬度比較侨颈，然后更新最大值即可余赢。

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) return 0;
        queue<pair<TreeNode*, int>> q;
        q.push(make_pair(root, 0));
        int max = INT_MIN;
        while(!q.empty()) {
            int size = q.size();
            int start = 0;
            int end;
            for (int i = 0; i < size; i++) {
                auto cur = q.front();
                q.pop();
                if (i == 0) {
                    start = cur.second;
                }
                end = cur.second;
                if (cur.first->left) {
                    q.push(make_pair(cur.first->left, cur.second * 2));
                }
                if (cur.first->right) {
                    q.push(make_pair(cur.first->right, cur.second * 2 + 1));
                }
            }
            max = max > (end - start + 1) ? max : end - start + 1;
        }
        return max;
    }
};