Same Tree
描述
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.
分析
无
递归版
- Java
- C++
// Same Tree
// 递归版,时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true; // 终止条件
if (p == null || q == null) return false; // 剪枝
return p.val == q.val // 三方合并
&& isSameTree(p.left, q.left)
&& isSameTree(p.right, q.right);
}
}
// Same Tree
// 递归版,时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
bool isSameTree(TreeNode *p, TreeNode *q) {
if (!p && !q) return true; // 终止条件
if (!p || !q) return false; // 剪枝
return p->val == q->val // 三方合并
&& isSameTree(p->left, q->left)
&& isSameTree(p->right, q->right);
}
};
迭代版
- Java
- C++
// Same Tree
// 迭代版,时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
Stack<TreeNode> s = new Stack<>();
s.push(p);
s.push(q);
while(!s.empty()) {
p = s.pop();
q = s.pop();
if (p == null && q == null) continue;
if (p == null || q == null) return false;
if (p.val != q.val) return false;
s.push(p.left);
s.push(q.left);
s.push(p.right);
s.push(q.right);
}
return true;
}
}
// Same Tree
// 迭代版,时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
bool isSameTree(TreeNode *p, TreeNode *q) {
stack<TreeNode*> s;
s.push(p);
s.push(q);
while(!s.empty()) {
p = s.top(); s.pop();
q = s.top(); s.pop();
if (!p && !q) continue;
if (!p || !q) return false;
if (p->val != q->val) return false;
s.push(p->left);
s.push(q->left);
s.push(p->right);
s.push(q->right);
}
return true;
}
};