任务是打印给定二叉树的左节点。首先，用户将插入数据，从而生成二叉树，然后打印所形成的树的左视图。
每个节点最多可以有 2 个子节点，因此这里程序必须仅遍历与节点关联的左指针
如果左指针不为空，则意味着它将有一些与之关联的数据或指针，否则它将是要打印并显示为输出的左子级。
示例Input : 1 0 3 2 4
Output : 1 0 2
这里，橙色节点代表二叉树的左视图。
在给定的图中，数据为 1 的节点是根节点，因此它将被打印，而不是转到左子节点，它将打印 0，然后它将转到 3 并打印其左子节点，即 2。
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我们可以使用递归方法来存储节点的级别并重复转移到
下面的代码显示了给定算法的 C 实现
算法
START
Step 1 -> create node variable of type structure
Declare int data
Declare pointer of type node using *left, *right
Step 2 -> create function for inserting node with parameter as new_data
Declare temp variable of node using malloc
Set temp->data = new_data
Set temp->left = temp->right = NULL
return temp
Step 3 -> declare function void left_view(struct node* root, int level, int* highest_level)
IF root = NULL
Exit
End
IF *highest_level < level
Print root->data
Set *highest_level = level
End
Recursively call left_view(root->left, level + 1, highest_level)
Recursively call left_view(root->right, level + 1, highest_level)
Step 4 -> Declare Function void left(struct node* root)
Set int highest_level = 0
Call left_view(root, 1, &highest_level)
Step 5-> In main()
Call New passing value user want to insert as struct node* root = New(1)
Call left(root)
STOP示例
#include#include //create a structure of a node struct node { int data; struct node *left, *right; //this pointer will point to the nodes attached with a node }; struct node* New(int new_data) { struct node* temp = (struct node*)malloc(sizeof(struct node)); //allocating memory to a pointer dynamically temp->data = new_data; temp->left = temp->right = NULL; return temp; } void left_view(struct node* root, int level, int* highest_level) { if (root == NULL) //if there is no node that means no data return; // this function will retrun the root node if there is only root node in a tree if (*highest_level < level) { printf("%d\t", root->data); *highest_level = level; } // Recursive function left_view(root->left, level + 1, highest_level); left_view(root->right, level + 1, highest_level); } void left(struct node* root) { int highest_level = 0; left_view(root, 1, &highest_level); } int main() { printf("left view of a binary tree is : "); struct node* root = New(1); root->left = New(0); root->right = New(3); root->right->left = New(2); root->right->right = New(4); left(root); return 0; }
输出
如果我们运行上面的程序，它将生成以下输出。
left view of a binary tree is : 1 0 2
