101. Symmetric Tree

subject ：

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100

Ideas 1：

Judge whether a tree is mirror symmetrical . First use bfs Here comes the stupid but easy to understand idea . On the basis of traversal by layer , Add a save Treenode* Of vector, Save all the children of the current layer , Including loopholes , And then put this vector Symmetrical comparison . The comparison is divided into five cases , Suppose the one on the left node by com[i], Dexter node by com[j]： First, left and right are empty , Then symmetrical ; Second, left space is not empty , Asymmetry ; Third, the left is not empty, the right is empty , Asymmetry ; Fourth, the left and right are not empty, but the values are not equal , Asymmetry ; Fifth, the left and right are not empty and the values are equal , Then symmetrical . As long as it is asymmetric , Go straight back to false that will do . This one more vector The solution is easy to understand , Is the complexity of taking up more space .

Code 1：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        while(q.size()) {
            int len = q.size();
            vector<TreeNode*> com;
            for (int i = 0; i < len; i++) {
                auto tmp = q.front();
                q.pop();
                com.push_back(tmp->left);
                com.push_back(tmp->right);
                if (tmp->left)
                    q.push(tmp->left);
                if (tmp->right)
                    q.push(tmp->right);
            }
            for (int i = 0, j = com.size() - 1; i < com.size() / 2; i++, j--) {
                if (!com[i] && !com[j])
                    continue;
                if (com[i] && com[j] && com[i]->val == com[j]->val)
                    continue;
                return false;
            }
        }
        return true;
    }
};

Ideas 2：

dfs How to write it , We know the five situations discussed above , That's it base case. Asymmetric situations return directly to false; And when the left and right are empty , We don't need to recurse down , Therefore, you can also directly return ture, The last case is that the left and right are not empty and the values are equal , We continue to recurse . We need to compare the outer and inner subtrees , That is, the left child of the left child should be the same as the right child of the right child , The right child of the left child is the same as the left child of the right child , Then return the results of both && that will do .

Code 2：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return dfs(root->left, root->right);
    }
private:
    bool dfs(TreeNode* left, TreeNode* right) {
        if (!left && !right)
            return true;
        else if (left && !right)
            return false;
        else if (!left && right)
            return false;
        else if (left->val != right->val)
            return false;
        bool waice = dfs(left->left, right->right);
        bool neice = dfs(left->right, right->left);
        return waice && neice;
    }
};