Leetcode—102.二叉树的层序遍历【中等】

news2024/11/22 14:20:26

2023每日刷题（二十四）

Leetcode—102.二叉树的层序遍历

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C语言BFS实现代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
#define MAXSIZE 2003

int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if(root == NULL) {
        return NULL;
    }
    *returnColumnSizes = (int *)malloc(sizeof(int) * MAXSIZE);
    int** ans = (int **)malloc(sizeof(int *) * MAXSIZE);
    struct TreeNode** queue = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * MAXSIZE);
    int front = 0, rear = 0;
    int len = 0;
    int pos = 0;
    queue[rear++] = root;
    while(front != rear) {
        len = rear - front;
        ans[*returnSize] = (int *)malloc(sizeof(int) * len);
        int cnt = 0;
        while(len > 0) {
            len--;
            struct TreeNode* tmp = queue[front++];
            ans[*returnSize][cnt++] = tmp->val;
            if(tmp->left) {
                queue[rear++] = tmp->left;
            }
            if(tmp->right) {
                queue[rear++] = tmp->right;
            }
        }
        (*returnColumnSizes)[*returnSize] = cnt;
        (*returnSize)++;
    }
    return ans;
}

运行结果

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C++BFS实现代码

/**
 * 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>> ans;
        if(root == NULL) {
            return ans;
        }
        TreeNode *p;
        queue<TreeNode*>qu;
        qu.push(root);
        vector<int>anslevel;
        while(!qu.empty()) {
            int n = qu.size();
            for(int i = 0; i < n; i++) {
                p = qu.front();
                qu.pop();
                anslevel.push_back(p->val);
                if(p->left != NULL) {
                    qu.push(p->left);
                }
                if(p->right != NULL) {
                    qu.push(p->right);
                }
            }
            ans.push_back(anslevel);
            anslevel.clear();
        }
        return ans;
    }
};