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。
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
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#include <stdio.h>
#include <stdlib.h>
//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;
}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
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