Maximum Width of Binary Tree问题及解法

问题描述：

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.

示例:

Input:            1         /   \        3     2       / \     \        5   3     9 Output: 4Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Input:           1         /          3           / \             5   3     Output: 2Explanation: The maximum width existing in the third level with the length 2 (5,3).

Input:           1         / \        3   2        /              5      Output: 2Explanation: The maximum width existing in the second level with the length 2 (3,2).

Input:           1         / \        3   2       /     \        5       9      /         \    6           7Output: 8Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

问题分析：

题目主要考查节点编号问题，对于一个左子节点，其编号为父节点编号的两倍，而右子节点编号为父节点编号两倍加1.

最终遍历每层找到最左最右节点编号，取其中最大的差值加1即为答案。

过程详见代码：

/** * 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 == nullptr) return 0;int res = 0;vector<pair<int, int>> val;val.emplace_back(pair<int, int>(1, 1));dfs(root->left, val, 1, 1, 0);dfs(root->right, val, 1, 1, 1);for (auto v : val){int  t = v.second - v.first + 1;if (t > res) res = t;}return res;}void dfs(TreeNode* root, vector<pair<int, int>>& val, int depth, int parent, int dirt){if (root == nullptr) return;if (val.size() <= depth) val.emplace_back(pair<int, int>(parent * 2 + dirt, parent * 2 + dirt));else val[depth].second = parent * 2 + dirt;dfs(root->left, val, depth + 1, parent * 2 + dirt, 0);dfs(root->right, val, depth + 1, parent * 2 + dirt, 1);}};