leetcode 662. Maximum Width of Binary Tree 深度优先遍历DFS

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).

那么其实只要我们知道了每一层中最左边和最右边的结点的位置，我们就可以算出这一层的宽度了。所以这道题的关键就是要记录每一层中最左边结点的位置，我们知道对于一棵完美二叉树，如果根结点是深度1，那么每一层的结点数就是2*n-1，那么每个结点的位置就是[1, 2*n-1]中的一个，假设某个结点的位置是i，那么其左右子结点的位置可以直接算出来，为2*i和2*i+1，可以自行带例子检验。由于之前说过，我们需要保存每一层的最左结点的位置，那么我们使用一个数组start，由于数组是从0开始的，我们就姑且认定根结点的深度为0，不影响结果。我们从根结点进入，深度为0，位置为1，进入递归函数。

首先判断，如果当前结点为空，那么直接返回，然后判断如果当前深度大于start数组的长度，说明当前到了新的一层的最左结点，我们将当前位置存入start数组中。然后我们用idx - start[h] + 1来更新结果res。这里idx是当前结点的位置，start[h]是当前层最左结点的位置。然后对左右子结点分别调用递归函数，注意左右子结点的位置可以直接计算出来，参见代码如下：

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    int widthOfBinaryTree(TreeNode* root)     {        int maxWidth = 0;        vector<int> start;        dfs(root, 0, 1, start, maxWidth);        return maxWidth;    }    void dfs(TreeNode* root, int depth, int index, vector<int>& start, int& maxWidth)    {        if (root == NULL)            return;        else        {            if (depth >= start.size())                start.push_back(index);            maxWidth = max(maxWidth, index - start[depth] + 1);            dfs(root->left, depth + 1, 2 * index, start, maxWidth);            dfs(root->right, depth + 1, 2 * index + 1, start, maxWidth);        }    }};