Data structures 如何将以下内容转换为二叉树?
将下列内容转换为二叉树的过程是什么?输出/答案是什么?A+(B+C+D+E)+F/G A+(B+C+D+E)+F/GData structures 如何将以下内容转换为二叉树?,data-structures,binary-tree,Data Structures,Binary Tree,将下列内容转换为二叉树的过程是什么?输出/答案是什么?A+(B+C+D+E)+F/G A+(B+C+D+E)+F/G result | \ A obj1 A+(B+C+D+E)+F/G result | \ A obj2 / \ obj1 F/G(could be linked to A for load balancing) result | \ A obj2 / \ obj4 obj5 | \ | \ obj3
result
| \
A obj1
A+(B+C+D+E)+F/G
result
| \
A obj2
/ \
obj1 F/G(could be linked to A for load balancing)
result
| \
A obj2
/ \
obj4 obj5
| \ | \
obj3 E F G
| \
obj1 D
/ \
B C
(B+C+D+E)
(B+C+D+E)
(B+C+D+E)
(B+C+D+E)
F/G
result
| \
A obj2
/ \
obj4 obj5
| \ |
obj3 E F
| \
obj1 D
/ \
B C
F/G
result
| \
A obj2
/ \
obj1 F/G(could be linked to A for load balancing)
result
| \
A obj2
/ \
obj4 obj5
| \ | \
obj3 E F G
| \
obj1 D
/ \
B C
树完成后,从下到上(所有叶子)开始计算,然后计算叶子完整的所有树枝
- B+C=obj1
- B+C+D=obj3
- B+C+D+E=obj4
- F/G=obj5
- (B+C+D+E)+F/G=obj2
- A+(B+C+D+E)+F/G=结果
result
| \
A obj2
/ \
obj4 F/G
| \
obj3 E
| \
obj1 D
/ \
B C
result
| \
A obj2
/ \
obj4 obj5
| \ |
obj3 E F
| \
obj1 D
/ \
B C
result
| \
A obj2
/ \
obj4 obj5
| \ | \
obj3 E F G
| \
obj1 D
/ \
B C