Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
1. 递归方式
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class Solution {
public:
bool (TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL)
return true;
else if (p == NULL && q != NULL || (p != NULL && q == NULL))
return false;
if (q->val != p->val)
return false;
if ( ! (isSameTree(p->left,q->left) && isSameTree(p->right,q->right)))
return false;
return true;
}
};
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2.非递归广度遍历
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class Solution {
public:
bool (TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL)
return true;
else if (p == NULL && q != NULL || (p != NULL && q == NULL))
return false;
queue<TreeNode *> pQueue;
queue<TreeNode *> qQueue;
pQueue.push(p);
qQueue.push(q);
bool result = true;
while ((! pQueue.empty()) && (! qQueue.empty()))
{
if (pQueue.size() != qQueue.size())
{
result = false;
break;
}
TreeNode *pNode = pQueue.front();
TreeNode *qNode = qQueue.front();
pQueue.pop();
qQueue.pop();
if ((pNode->val != qNode->val)
|| (pNode->left == NULL && qNode->left != NULL)
|| (pNode->left != NULL && qNode->left == NULL)
|| (pNode->right == NULL && qNode->right != NULL)
|| (pNode->right != NULL && qNode->right == NULL))
{
result = false;
break;
}
if (pNode->left != NULL)
pQueue.push(pNode->left);
if (qNode->left != NULL)
qQueue.push(qNode->left);
if (pNode->right != NULL)
pQueue.push(pNode->right);
if (qNode->right != NULL)
qQueue.push(qNode->right);
}
if (result && (!pQueue.empty() || !qQueue.empty()))
result = false;
return result;
}
};
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