The sword refers to the second edition of the offer interview question 28: symmetric binary tree (java)

Topic description:
Please implement a function to determine whether a binary tree is symmetric. Note that a binary tree is symmetric if it is the same as its mirror image.

Analysis:
For a binary tree, traverse from the root node,
if one of the left and right child nodes is NULL, then it is definitely not a symmetric binary tree;
if the left and right child nodes are not empty, but not equal, then it is definitely not a symmetric binary tree;
If the left and right child nodes are not empty and equal, then
traverse left subtree, the traversal order is: current node, left subtree, right subtree;
traverse the right subtree, the traversal order is: current node, right subtree , the left subtree;
if the sequence of traversing the left subtree is the same as the sequence of traversing the right subtree, then the binary tree is a symmetric binary tree. (recursive implementation)

code show as below:

/**
 * 对称的二叉树
 */
public class SymmetricalBinaryTree {

    public boolean isSymmetrical(TreeNode root) {
        return isSymmetrical(root,root);
    }

    public boolean isSymmetrical(TreeNode root1,TreeNode root2){
        //如果两个根节点都是null
        if(root1 == null&&root2 == null){
            return true;
        }

        //如果只有一个根节点是null
        if(root1 == null || root2 == null){
            return false;
        }

        /**
         * 两个根节点都不为null
         */
        if(root1.val != root2.val){
            return false;
        }

        return isSymmetrical(root1.left, root2.right) && isSymmetrical(root1.right,root2.left);
    }

}

//二叉树的定义
class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}