Breadth-first Search -- Leetcode problem515. Find Largest Value in Each Tree Row

• 描述：You need to find the largest value in each row of a binary tree.
  Example:

Input:

      1     / \    3   2   / \   \    5   3   9 

Output: [1, 3, 9]

• 分析：这道题要求找到每层树中最大的元素，并输出。

• 思路一：用bfs对整棵数进行搜索，并寻找最大值。我采用了set内置的排序对每层的元素进行排序，将每次得到的set的最后一个元素加入vector。

/** * 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:   vector<int> largestValues(TreeNode* root) {    vector<int> result;    if (!root) return result;    set<int> my_set;    set<int>::iterator it;    queue<TreeNode*> my_queue;    result.push_back(root -> val);    my_queue.push(root);    while (!my_queue.empty()) {        int n = my_queue.size();        my_set.clear();        for (int i = 0; i < n; i ++) {            TreeNode* temp = my_queue.front();            my_queue.pop();            if (temp -> left) {                my_set.insert(temp -> left -> val);                my_queue.push(temp -> left);            }            if (temp -> right) {                my_set.insert(temp -> right -> val);                my_queue.push(temp -> right);            }        }        if (!my_set.empty()) {            it = my_set.end();            it--;            result.push_back(*it);        }    }    return result;}};

• 思路二：用递归的方式实现。（dfs）

class Solution {public:   void dfs(TreeNode* root, vector<int>& result, int deepth) {    if (!root) return;    if (deepth > result.size()) result.push_back(root -> val);    else result[deepth - 1] = max(result[deepth - 1], root -> val);    dfs(root -> left, result, deepth + 1);    dfs(root -> right,result, deepth + 1);}vector<int> largestValues(TreeNode* root) {    vector<int> result;    if (!root) return result;    dfs(root, result, 1);    return result;}};