C语言刷二叉树（二）基础部分

102. 二叉树的层序遍历

/**
 * 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().
 */
int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes)
{
    struct TreeNode *queue[2001];
    int front = 0, rear = 0;

    if (root != NULL) {
        queue[rear++] = root;
    }

    int **res = (int **)malloc(sizeof(int *) * 2001);
    *returnColumnSizes = (int *)malloc(sizeof(int) * 2001);
    *returnSize = 0;

    while (front < rear) {
        int colIndex = 0;
        // 申请内存
        int last = rear;
        res[(*returnSize)] = (int *)malloc(sizeof(int) * (last - front));
        while (front < last) {
            // 取出当前节点
            struct TreeNode *curNode = queue[front++];
            res[(*returnSize)][colIndex++] = curNode->val;

            if (curNode->left != NULL) {
                queue[rear++] = curNode->left;
            }
            if (curNode->right != NULL) {
                queue[rear++] = curNode->right;
            }
        }
        (*returnColumnSizes)[(*returnSize)] = colIndex;
        (*returnSize)++;
    }
    return res;
}

226. 翻转二叉树

这道题注意：写指针的Swap的时候，要取指针的地址，即二级指针

方法1：递归（前序遍历）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

void Swap(struct TreeNode **left, struct TreeNode **right)
{
    // printf("begin left:%p, right:%p\n", left, right);
    struct TreeNode *tmp = *left;
    *left = *right;
    *right = tmp;
    // printf("end left:%p, right:%p\n", left, right);
}

struct TreeNode* invertTree(struct TreeNode* root)
{
    if (root == NULL) {
        return root;
    }
    // 前序：中 左 右
    // printf("Swap前 end left:%p, right:%p\n", root->left, root->right);
    Swap(&root->left, &root->right);
    // printf("Swap后 end left:%p, right:%p\n", root->left, root->right);
    // struct TreeNode* temp = root->left;
    // root->left = root->right;
    // root->right = temp;

    invertTree(root->left);
    invertTree(root->right);
    return root;
}

方法2：迭代（前序遍历）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

void Swap(struct TreeNode **left, struct TreeNode **right)
{
    struct TreeNode *tmp = *left;
    *left = *right;
    *right = tmp;
}

struct TreeNode* invertTree(struct TreeNode* root)
{
    struct TreeNode *stack[10];
    int index = 0;

    if (root != NULL) {
        stack[index++] = root;
    }
    while (index != 0) {
        // 取出当前节点（准备交换左右子节点）
        struct TreeNode *curNode = stack[--index];
        if (curNode == NULL) {
            break;
        }
        Swap(&curNode->left, &curNode->right);
        // 上面已经交换完了，这里安装前序遍历正常顺序（如果不是翻转的需要则逆序，由于是栈）
        if (curNode->left != NULL) {
            stack[index++] = curNode->left;
        }
        if (curNode->right != NULL) {
            stack[index++] = curNode->right;
        }
    }
    return root;
}