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);
        }
    }
};