题解 | #对称的二叉树#

1. 先判断根节点不存在，返回true
2. 递归调用，判断3种情况
3. 返回

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }

}
*/
public class Solution {
    boolean isSymmetrical(TreeNode pRoot) {
        //根节点不存在，返回true
        if(pRoot == null){
            return true;
        }
        return symmetricTree(pRoot.left,pRoot.right);
    }
    //递归
    public static boolean symmetricTree(TreeNode l,TreeNode r){
        //都为null 返回true
        if(l == null && r==null){
            return true;
        }
        //二者不一致 返回false
        if(l !=null ^ r !=null){
            return false;
        }
        //二者都存在，值不等返回false
        if(l.val != r.val){
            return false;
        }
        return symmetricTree(l.left,r.right) && symmetricTree(l.right,r.left);
    }
}