题目描述:
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 scould 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.
求解方法:
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool doesTree1hasTree2(TreeNode* tree1, TreeNode* tree2)
{
if(tree2 == NULL && tree1 == NULL) return true;
if(tree2 == NULL || tree1 == NULL) return false;
if(tree1->val == tree2->val)
{
return doesTree1hasTree2(tree1->left,tree2->left) && doesTree1hasTree2(tree1->right, tree2->right);
}
else
{
return false;
}
}
bool isSubtree(TreeNode* s, TreeNode* t)
{
bool res = false;
if(t == NULL) res = true;
if(s == NULL) res = false;
if(s && t)
{
if(s->val == t->val)
res = doesTree1hasTree2(s,t);
if(res == false)
{
res = isSubtree(s->left,t);
}
if(res == false)
{
res = isSubtree(s->right,t);
}
}
return res;
}
};
小结:写这种递归问题时,最关键的是要搞清楚终止条件,使其可以满足不同情况,要不然很有可能在各种特殊case下出错。