LeetCode | 二叉树的前中后序遍历

LeetCode | 二叉树的前中后序遍历

OJ链接

• 这里我们使用递归的方法来解决
• 这里题目还要求我们返回这棵树的根
• 我们这里需要先算出这个树有多大
• 然后开辟空间
• 再进行前序的遍历

void preorder(struct TreeNode* root,int* a,int* pi)
{
    if(root == NULL)
        return;
    a[(*pi)++] = root->val;

    preorder(root->left,a,pi);
    preorder(root->right,a,pi);
}
int TreeSize(struct TreeNode* root)
{
    return root == NULL ? 0 : TreeSize(root->left) + TreeSize(root->right) + 1;
}

int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    //计算树有多少个节点
    int n = TreeSize(root);
    *returnSize = n;
    //开辟n个大小
    int* a = malloc(sizeof(int) * n);

    int i = 0;
    //前序遍历
    preorder(root,a,&i);
    return a;
}

• 这里前序遍历完成后，我们的中序和后序也是一样的，直接CV即可

• 中序遍历：OJ链接

int TreeSize(struct TreeNode* root)
{
    return root == NULL ? 0 : TreeSize(root->left) + TreeSize(root->right) + 1;
}

void inorder(struct TreeNode* root,int* a ,int* pi)
{
    if(root == NULL)
        return;
    
    inorder(root->left,a,pi);
    a[(*pi)++] = root->val;
    inorder(root->right,a,pi);
}

int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    int n = TreeSize(root);
    int* a = (int*)malloc(sizeof(int) * n);

    *returnSize = n;
    int i = 0;
    inorder(root,a,&i);

    return a;
}

• 后序遍历：OJ链接

int TreeSize(struct TreeNode* root)
{
    return root == NULL ? 0 : TreeSize(root->left) + TreeSize(root->right) + 1;
}

void postorder(struct TreeNode* root,int* a ,int* pi)
{
    if(root == NULL)
        return;
    
    postorder(root->left,a,pi);
    postorder(root->right,a,pi);
    a[(*pi)++] = root->val;

}
int* postorderTraversal(struct TreeNode* root, int* returnSize) {
    int n = TreeSize(root);
    int* a = (int*)malloc(sizeof(int) * n);

    *returnSize = n;
    int i = 0;
    postorder(root,a,&i);

    return a;
}