100.Same Tree
Given the roots of two binary trees p and q, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.
Example 1:
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| Input: p = [1,2,3], q = [1,2,3]
Output: true
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Example 2:
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| Input: p = [1,2], q = [1,null,2]
Output: false
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Example 3:
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| Input: p = [1,2,1], q = [1,1,2]
Output: false
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深度优先搜索：
时间复杂度：O(min(m,n))
空间复杂度：O(min(m,n))
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| bool isSameTree(TreeNode* p, TreeNode* q) {
if (p == nullptr && q == nullptr) {
return true;
} else if (p == nullptr || q == nullptr) {
return false;
} else if (p->val != q->val) {
return false;
} else {
return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
}
}
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广度优先搜索：
时间复杂度：O(min(m,n))
空间复杂度：O(min(m,n))
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| bool isSameTree(TreeNode* p, TreeNode* q) {
if (p == nullptr && q == nullptr) {
return true;
} else if (p == nullptr || q == nullptr) {
return false;
}
queue <TreeNode*> queue1, queue2;
queue1.push(p);
queue2.push(q);
while (!queue1.empty() && !queue2.empty()) {
auto node1 = queue1.front();
queue1.pop();
auto node2 = queue2.front();
queue2.pop();
if (node1->val != node2->val) {
return false;
}
auto left1 = node1->left, right1 = node1->right, left2 = node2->left, right2 = node2->right;
if ((left1 == nullptr) ^ (left2 == nullptr)) {
return false;
}
if ((right1 == nullptr) ^ (right2 == nullptr)) {
return false;
}
if (left1 != nullptr) {
queue1.push(left1);
}
if (right1 != nullptr) {
queue1.push(right1);
}
if (left2 != nullptr) {
queue2.push(left2);
}
if (right2 != nullptr) {
queue2.push(right2);
}
}
return queue1.empty() && queue2.empty();
}
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