408算法题leetcode--第16天

144. 二叉树的前序遍历

• 144. 二叉树的前序遍历
• 思路：递归和非递归
• 时间：O(n)；空间：O(n)

/**
 * 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:
    vector<int>ret;

    void pre(TreeNode* root){
        if(root != nullptr){
            ret.push_back(root->val);
            pre(root->left);
            pre(root->right);
        }
    }
    vector<int> preorderTraversal(TreeNode* root) {
        pre(root);
        return ret;
    }
};

102. 二叉树的层序遍历

• 102. 二叉树的层序遍历
• 思路：bfs，队列
• 时间：O(n)；空间：O(n)

/**
 * 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:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>>ret;
        if(!root) return ret;

        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty()){
            vector<int>level;
            int size = q.size();
            for(int i = 0; i < size; i++){
                TreeNode* temp = q.front();
                q.pop();
                level.push_back(temp->val);
                if(temp->left) q.push(temp->left);
                if(temp->right) q.push(temp->right);
            }
            ret.push_back(level);
        }
        return ret;
    }
};

199. 二叉树的右视图

• 199. 二叉树的右视图
• 思路和时间，空间：同上

/**
 * 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:
    vector<int> rightSideView(TreeNode* root) {
        vector<int>ret;
        if(!root) return ret;
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; i++){
                TreeNode* temp = q.front();
                q.pop();
                if(i == size - 1){
                    ret.push_back(temp->val);
                }
                if(temp->left) q.push(temp->left);
                if(temp->right) q.push(temp->right);
            }
        }
        return ret;
    }
};

404. 左叶子之和

• 404. 左叶子之和
• 思路：dfs
• 时间和空间：O(n)

/**
 * 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:
    int traverse(TreeNode* root, bool is_left){
        if(!root) return 0;  // 节点为空
        if(!root->left && !root->right && is_left) return root->val;  // 叶子节点且为左子树 
        return traverse(root->left, true) + traverse(root->right, false);
    }
    int sumOfLeftLeaves(TreeNode* root) {
        // 三种情况：1. 节点为空；2. 叶子节点且为左子树 ; 3. else
        return traverse(root, false);
    }
};

104. 二叉树的最大深度

• 104. 二叉树的最大深度
• 思路和时间空间：同上

/**
 * 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:
    int maxDepth(TreeNode* root) {
        // 1. 有左右儿子，2. 空, 3. else
        if(!root) return 0;
        int max_v = 1;
        if(root->left) max_v = maxDepth(root->left) + 1;
        if(root->right) max_v = max(max_v, maxDepth(root->right) + 1);
        return max_v;
    }
};

原文地址：https://blog.csdn.net/weixin_58073817/article/details/142557688

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