[Leetcode] 100. Same Tree 解题报告
2017-04-21 10:27
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题目:
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.
思路:
Easy级别的题目,没什么可说的,直接用递归。时间复杂度是O(n),空间复杂度取决于树是否是平衡的,如果是平衡的,则为O(logn),否则最高可以达到O(n)。
代码:
/**
* 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 isSameTree(TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL) {
return true;
}
else if (p == NULL) {
return false;
}
else if (q == NULL) {
return false;
}
else {
if (p->val != q->val) {
return false;
}
if (!isSameTree(p->left, q->left)) {
return false;
}
if (!isSameTree(p->right, q->right)) {
return false;
}
return true;
}
}
};
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.
思路:
Easy级别的题目,没什么可说的,直接用递归。时间复杂度是O(n),空间复杂度取决于树是否是平衡的,如果是平衡的,则为O(logn),否则最高可以达到O(n)。
代码:
/**
* 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 isSameTree(TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL) {
return true;
}
else if (p == NULL) {
return false;
}
else if (q == NULL) {
return false;
}
else {
if (p->val != q->val) {
return false;
}
if (!isSameTree(p->left, q->left)) {
return false;
}
if (!isSameTree(p->right, q->right)) {
return false;
}
return true;
}
}
};
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