【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
572. Subtree of Another Tree
Given two non-empty binary trees
s
and
t
, check whether tree
t
has exactly the same structure and node values with a subtree of
s
. A subtree of
s is a tree consists of a node in s
and all of this node's descendants. The tree
s
could also be considered as a subtree of itself.
Example 1:
Given tree s:
3
/ \
4 5
/ \
1 2
Given tree t:
4
/ \
1 2
Return
true
, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:
3
/ \
4 5
/ \
1 2
/
0
Given tree t:
4
/ \
1 2
Return
false
.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
问题：判断某子树是否是主树的子结构
方法：前序遍历的递归方法
主树s， 子树t
*/
class
Solution
{
public
:
bool
isSubtree
(
TreeNode
*
s
,
TreeNode
*
t
)
//遍历主树s（前序遍历递归法）
{
if(s == nullptr)
return
false
;
//前序遍历递归的出口，
注意一定要加判断空指针的语句，后面有s->left和s->right
//前序遍历（递归法）主树t各结点,从根结点到左子树再到右子树（s一直在延伸，展开分支）
if
(
isSame
(
s
,
t
))
//判断当前结点下的子树是否一样
return
true
;
else
//判断当前结点的左子树是否为子树t,再判断右子树是否为子树t
return
isSubtree
(
s
->
left
,
t
)
||
isSubtree
(
s
->
right
,
t
);
}
private
:
bool
isSame
(
TreeNode
*
s
,
TreeNode
*
t
)
//确定s父结点后，开始同时扫描（递归法）s和t，看是各个结点是否相等
{
if
(
s
==
nullptr
&&
t
==
nullptr
)
return
true
;
//如果最后均遍历到空结点，返回true
else
if
(
s
==
nullptr
||
t
==
nullptr
)
return
false
;
//如果一个遍历到空，一个没有，说明不同，返回false
if
(
s->val == t->val
)
{
return
isSame(s->left, t->left) && isSame(s->right, t->right);
}
else
return
false
;
//不相同，返回假
}
};