[Leetcode -94. In-order traversal of binary trees -145. Post-order traversal of binary trees]

Leetcode -94. In-order traversal of binary tree

Question : Given the root node root of a binary tree, return its in-order traversal.

Example 1:
Input: root = [1, null, 2, 3]
Output: [1, 3, 2]

Example 2:
Input: root = []
Output: []

Example 3:
Input: root = [1]
Output: [1]

Tip:
The number of nodes in the tree is in the range [0, 100]

• 100 <= Node.val <= 100

Idea : In-order traversal of a binary tree is turned into a sub-problem: first traverse the left subtree of the current root, then print the value of the current root, and finally traverse the right subtree of the current root;

		void Inorder(struct TreeNode* root, int* a, int* pos)
		{
		    if (root == NULL)
		        return;
		
		    //先递归当前根的左子树；再将当前根的 val 存放到数组中；最后递归当前根的右子树
		    Inorder(root->left, a, pos);
		    a[(*pos)++] = root->val;
		    Inorder(root->right, a, pos);
		}
		
		
		
		int* inorderTraversal(struct TreeNode* root, int* returnSize)
		{
		    //开辟一个返回中序遍历的数组，pos记录数组的长度
		    int* ret = (int*)malloc(sizeof(int) * 100);
		    int pos = 0;
		
		    //进入中序遍历
		    Inorder(root, ret, &pos);
		    *returnSize = pos;
		    return ret;
		}

Leetcode -145. Post-order traversal of binary tree

Question : Given the root node root of a binary tree, return the post-order traversal of its node value.

Example 1:
Input: root = [1, null, 2, 3]
Output: [3, 2, 1]

Example 2:
Input: root = []
Output: []

Example 3:
Input: root = [1]
Output: [1]

Tip:
The number of nodes in the tree is in the range [0, 100]

• 100 <= Node.val <= 100

Idea : Post-order traversal of a binary tree is turned into a sub-problem: first traverse the left subtree of the current root, then traverse the right subtree of the current root, and finally print the value of the current root;

		void Postorder(struct TreeNode* root, int* a, int* pos)
		{
		    if (root == NULL)
		        return;
		
		    //先递归当前根的左子树；再递归当前根的右子树；最后将当前根的 val 存放到数组中
		    Postorder(root->left, a, pos);
		    Postorder(root->right, a, pos);
		    a[(*pos)++] = root->val;
		}
		
		
		int* postorderTraversal(struct TreeNode* root, int* returnSize)
		{
		    //开辟返回的数组
		    int* ret = (int*)malloc(sizeof(int) * 100);
		    int pos = 0;
		
		    //进入后序遍历
		    Postorder(root, ret, &pos);
		    *returnSize = pos;
		    return ret;
		}