Problem: 515. 在每个树行中找最大值

题目描述

思路

思路1：BFS

套用BFS模板，直接在遍历树的某一层时将当前层的最大值存入数组中

思路2：DFS

回溯思想，在递归时不断更新可选列表（根据当前树的层数，也可以抽象看作是回溯思想中的决策阶段）

复杂度

思路1：
时间复杂度:

$O(n)$；其中$n$为二叉树节点的个数

空间复杂度:

$O(n)$

思路2：
时间复杂度:

$O(n)$

空间复杂度:

$O(height)$；其中$height$为二叉树的高度

Code

思路1：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:/*** Get the largest values of each level** @param root The root of a binary tree* @return vector<int>*/vector<int> largestValues(TreeNode* root) {if (root == nullptr) {return{};}return levelMaxNumber(root);}/*** BFS* * @param root The root of a binary tree* @return vector<int>*/vector<int> levelMaxNumber(TreeNode* root) {vector<int>res;queue<TreeNode*> queue;int depth = 0;queue.push(root);res.push_back(root -> val);while (!queue.empty()) {int levelMaxNum = INT_MIN;int curLevelSize = queue.size();for (int i = 0; i < curLevelSize; ++i) {TreeNode* curLevelNode = queue.front();queue.pop();if (curLevelNode -> left != nullptr) {TreeNode* nextLeftNode = curLevelNode -> left;queue.push(nextLeftNode);levelMaxNum = max(nextLeftNode -> val, levelMaxNum);}if (curLevelNode -> right != nullptr) {TreeNode* nextRightNode = curLevelNode -> right;queue.push(nextRightNode);levelMaxNum = max(nextRightNode -> val, levelMaxNum);}}if (!queue.empty()) {res.push_back(levelMaxNum);}}return res;}
};

思路2：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:/*** Get the largest values of each level** @param root The root of a binary tree* @return vector<int>*/vector<int> largestValues(TreeNode *root) {vector<int> res;if (root == nullptr) {return {};}dfs(res, root, 0);return res;}/*** DFS** @param res Result set* @param root The root of a binary tree* @param curHeight The current height of a binary tree*/void dfs(vector<int> &res, TreeNode *root, int curHeight) {if (curHeight == res.size()) {res.push_back(root->val);} else {res[curHeight] = max(res[curHeight], root->val);}if (root->left) {dfs(res, root->left, curHeight + 1);}if (root->right) {dfs(res, root->right, curHeight + 1);}}
};