LeetCode | Pre-middle and post-order traversal of binary tree
- Here we use recursive method to solve
- The question here also requires us to return to the root of the tree
- Here we need to first calculate how big the tree is
- Then make space
- Then perform pre-order traversal
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;
}
-
After the pre-order traversal is completed here, our in-order and post-order are also the same, just do CV directly.
-
In-order traversal:OJ link
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;
}
- Post-order traversal:OJ link
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;
}