题解 | #对称的二叉树#TOP31

思路:
1.递归，遍历左子树和右子树对比
2.终止条件，如果当前两个要遍历的节点为空。递归条件，当前两个节点的值相等，就继续递归下去。

/*
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) {
        if(pRoot == null){
            return true;
        }
        return isSame(pRoot.left, pRoot.right);
    }
    boolean isSame(TreeNode node1,TreeNode node2){
        if(node1 == null && node2 == null){
            return true;
        }
        if(node1 == null || node2 == null){
            return false;
        }
        if(node1.val == node2.val){
             return isSame(node1.left, node2.right) && isSame(node1.right, node2.left);
        }
        return false;
    }
}