leetcode 662. Maximum Width of Binary Tree 二叉树最大宽度 + 深度优先遍历DFS 我就是我 2022-06-03 07:22 108阅读 0赞 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); } } };
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