二叉表达式树是一种表达算术表达式的二叉树。二叉表达式树中的每一个节点都有零个或两个子节点。 叶节点（有 0 个子节点的节点）表示操作数，非叶节点（有 2 个子节点的节点）表示运算符。在本题中，我们只考虑 '+' 运算符（即加法）。
给定两棵二叉表达式树的根节点 root1 和 root2 。如果两棵二叉表达式树等价，返回 true ，否则返回 false 。
当两棵二叉搜索树中的变量取任意值，分别求得的值都相等时，我们称这两棵二叉表达式树是等价的。
示例 1:
输入： root1 = [x], root2 = [x] 输出： true
示例 2:
输入：root1 = [+,a,+,null,null,b,c], root2 = [+,+,a,b,c]
输出：true
解释：a + (b + c) == (b + c) + a
示例 3:
输入： root1 = [+,a,+,null,null,b,c], root2 = [+,+,a,b,d]
输出： false
解释： a + (b + c) != (b + d) + a
提示：
- 两棵树中的节点个数相等，且节点个数为范围
[1, 4999]内的奇数。 Node.val是'+'或小写英文字母。- 给定的树保证是有效的二叉表达式树。
进阶：当你的答案需同时支持 '-' 运算符（减法）时，你该如何修改你的答案
# Definition for a binary tree node.
# class Node(object):
# def __init__(self, val=" ", left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def checkEquivalence(self, root1: 'Node', root2: 'Node') -> bool:
counter = [0] * 26
def dfs(root, incr):
if root:
dfs(root.left, incr)
dfs(root.right, incr)
if root.val != '+':
counter[ord(root.val) - ord('a')] += incr
dfs(root1, 1)
dfs(root2, -1)
return counter.count(0) == 26# Definition for a binary tree node.
# class Node(object):
# def __init__(self, val=" ", left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def checkEquivalence(self, root1: 'Node', root2: 'Node') -> bool:
def calc(ans, left, right, op):
for i in range(26):
if op == '+':
ans[i] = left[i] + right[i]
else:
ans[i] = left[i] - right[i]
def dfs(root):
ans = [0] * 26
if not root:
return ans
if root.val in ['+', '-']:
left, right = dfs(root.left), dfs(root.right)
calc(ans, left, right, root.val)
else:
ans[ord(root.val) - ord('a')] += 1
return ans
return dfs(root1) == dfs(root2)/**
* Definition for a binary tree node.
* class Node {
* char val;
* Node left;
* Node right;
* Node() {this.val = ' ';}
* Node(char val) { this.val = val; }
* Node(char val, Node left, Node right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int[] counter;
public boolean checkEquivalence(Node root1, Node root2) {
counter = new int[26];
dfs(root1, 1);
dfs(root2, -1);
for (int n : counter) {
if (n != 0) {
return false;
}
}
return true;
}
private void dfs(Node root, int incr) {
if (root == null) {
return;
}
dfs(root.left, incr);
dfs(root.right, incr);
if (root.val != '+') {
counter[root.val - 'a'] += incr;
}
}
}/**
* Definition for a binary tree node.
* class Node {
* char val;
* Node left;
* Node right;
* Node() {this.val = ' ';}
* Node(char val) { this.val = val; }
* Node(char val, Node left, Node right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean checkEquivalence(Node root1, Node root2) {
int[] ans1 = dfs(root1);
int[] ans2 = dfs(root2);
for (int i = 0; i < 26; ++i) {
if (ans1[i] != ans2[i]) {
return false;
}
}
return true;
}
private int[] dfs(Node root) {
int[] ans = new int[26];
if (root == null) {
return ans;
}
if (root.val == '+' || root.val == '-') {
int[] left = dfs(root.left);
int[] right = dfs(root.right);
calc(ans, left, right, root.val);
} else {
++ans[root.val - 'a'];
}
return ans;
}
private void calc(int[] ans, int[] left, int[] right, char op) {
for (int i = 0; i < 26; ++i) {
ans[i] = op == '+' ? left[i] + right[i] : left[i] - right[i];
}
}
}/**
* Definition for a binary tree node.
* struct Node {
* char val;
* Node *left;
* Node *right;
* Node() : val(' '), left(nullptr), right(nullptr) {}
* Node(char x) : val(x), left(nullptr), right(nullptr) {}
* Node(char x, Node *left, Node *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> counter;
bool checkEquivalence(Node* root1, Node* root2) {
counter.resize(26);
dfs(root1, 1);
dfs(root2, -1);
return count(counter.begin(), counter.end(), 0) == 26;
}
void dfs(Node* root, int incr) {
if (!root) return;
dfs(root->left, incr);
dfs(root->right, incr);
if (root->val != '+') counter[root->val - 'a'] += incr;
}
};/**
* Definition for a binary tree node.
* struct Node {
* char val;
* Node *left;
* Node *right;
* Node() : val(' '), left(nullptr), right(nullptr) {}
* Node(char x) : val(x), left(nullptr), right(nullptr) {}
* Node(char x, Node *left, Node *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool checkEquivalence(Node* root1, Node* root2) {
return dfs(root1) == dfs(root2);
}
vector<int> dfs(Node* root) {
vector<int> ans(26);
if (!root) return ans;
if (root->val == '+' || root->val == '-')
{
auto left = dfs(root->left);
auto right = dfs(root->right);
calc(ans, left, right, root->val);
return ans;
}
++ans[root->val - 'a'];
return ans;
}
void calc(vector<int>& ans, vector<int>& left, vector<int>& right, char op) {
for (int i = 0; i < 26; ++i)
ans[i] = op == '+' ? left[i] + right[i] : left[i] - right[i];
}
};