给你一个二叉树的根节点 root , 检查它是否轴对称。
示例 1:
输入:root = [1,2,2,3,4,4,3] 输出:true
示例 2:
输入:root = [1,2,2,null,3,null,3] 输出:false
提示:
树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100
进阶:你可以运用递归和迭代两种方法解决这个问题吗?
递归法:
class Solution {
public:
bool compare(TreeNode* left,TreeNode* right){
//排除空节点情况
if(left!=NULL && right==NULL)return false;
else if(left==NULL && right!=NULL)return false;
else if(left==NULL && right==NULL)return true;
//排除内容是否一致
else if(left->val!=right->val)return false;
//此时内容一致,做递归判断下一层
bool outside=compare(left->left,right->right);
bool inside=compare(left->right,right->left);
bool isSame=outside && inside;
return isSame;
}
bool isSymmetric(TreeNode* root) {
if(root==NULL)return true;
return compare(root->left,root->right);
}
};
递归遍历化简版:
class Solution {
public:
bool compare(TreeNode* left,TreeNode* right){
if(left!=NULL&&right==NULL)return false;
else if(left==NULL&&right!=NULL)return false;
else if(left==NULL&&right==NULL)return true;
else if(left->val!=right->val)return false;
else return compare(left->left,right->right)&&compare(left->right,right->left);
}
bool isSymmetric(TreeNode* root) {
if(root==NULL)return true;
return compare(root->left,root->right);
}
};
使用队列:
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(root==NULL)return true;
queueque;
que.push(root->left);
que.push(root->right);
while(!que.empty()){
TreeNode* leftNode=que.front();que.pop();
TreeNode* rightNode=que.front();que.pop();
if(!leftNode&&!rightNode)continue;
if(!leftNode||!rightNode||(leftNode->val!=rightNode->val))return false;
que.push(leftNode->left);
que.push(rightNode->right);
que.push(leftNode->right);
que.push(rightNode->left);
}
return true;
}
};
使用栈:
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(root==NULL)return true;
stackst;
st.push(root->left);
st.push(root->right);
while(!st.empty()){
TreeNode* leftNode=st.top();st.pop();
TreeNode* rightNode=st.top();st.pop();
if(!leftNode&&!rightNode)continue;
if(!leftNode||!rightNode||(leftNode->val!=rightNode->val))return false;
st.push(leftNode->left);
st.push(rightNode->right);
st.push(leftNode->right);
st.push(rightNode->left);
}
return true;
}
};



