Symmetric Binary Tree-Simple Tree Traversal

Symmetric binary tree

Please implement a function to determine whether a binary tree is symmetric. If a binary tree is the same as its mirror image, then it is symmetric.

For example, the binary tree [1,2,2,3,4,4,3] is symmetric.

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not mirror symmetry:

    1
   / \
  2   2
   \   \
   3    3

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

Limit:
0 <= number of nodes <= 1000

---

analysis

Simple tree traversal

solution

A tree is divided into two trees and compared on the premise that their root nodes are the same: the left node of the A tree and the right node of the B tree; the right node of the A tree and the left node of the B tree;

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

bool judge(struct TreeNode* A, struct TreeNode * B){
    
    
    // 如果A B都为空了，得到true值，
    if(A == NULL && B == NULL)
        return true;
    // 在AB同空之后，判断A或者B是否有一方为空，如果是，则返回false，确保程序的鲁棒性，否则继续往下执行则会报错，空节点没有相关属性。
    if(A == NULL || B == NULL)
        return false;
    // A与B的关系，如果没有上诉的递归出口，下面这段程序将会陷入死循环。
    if(A->val != B->val)
        return false;
    else{
    
    
        return judge(A->left, B->right) && judge(A->right, B->left);
    }
}

bool isSymmetric(struct TreeNode* root){
    
    

    if(root == NULL){
    
    
        return true;
    }
    struct TreeNode * p = root->right;
    struct TreeNode * q = root->left;
    return judge(q, p);
}