【Sword Finger 28】Symmetric Binary Tree

Method 1: Whether the left and right subtrees are symmetric, time O(n), space O(n)

Solution: For two symmetrical nodes, there are

1. node1->val == node2->val, the two nodes have the same value
2. node1->left->val == node1->right->val, the left tree of node1 is the same as the right tree of node2
3. node1->right->val == node2->left->val, the right tree of node1 is the same as the left tree of node2
   

/**
 * 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 isSymmetric(TreeNode* root) 
    {
    
    
        // 1.判断树的左右树是否对称--》判断两颗树是否相同
        // 2.先写出判断两个树是否相同的代码
        // 3.然后修改条件变成判断对称
        if (root == nullptr)
            return true;
        return isSyTree(root->left, root->right);
    }
    bool isSyTree(TreeNode* node1, TreeNode* node2)
    {
    
    
        if (node1 == nullptr && node2 == nullptr)
            return true;
        if (node1 == nullptr || node2 == nullptr || node1->val != node2->val)
            return false;
        return isSyTree(node1->left, node2->right) && isSyTree(node1->right, node2->left);
    }
};