题目描述
分析
1. 递归
将头节点的左右子树拆分为两个数，同时先序遍历左子树和镜像先序遍历右子树，对比数值
2. 层级遍历
todo ,层级遍历每一层，记录到list中，空节点也要记录，最后对每一层的左右两侧节点数值对比
代码实现
public enum Q101 {
instance;
// 递归
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
return isSymmetric(root.left,root.right);
}
private boolean isSymmetric(TreeNode left, TreeNode right) {
if(left!=null&&right!=null){
if(left.val==right.val){
return isSymmetric(left.left,right.right)&&isSymmetric(left.right,right.left);
}else {
return false;
}
}else if(left==null&&right==null){
return true;
}else {
return false;
}
}
public static void main(String[] args) {
TreeNode root1=new TreeNode(1);
root1.left=new TreeNode(2);
root1.right=new TreeNode(2);
root1.left.left=new TreeNode(3);
root1.left.right=new TreeNode(4);
root1.right.left=new TreeNode(4);
root1.right.right=new TreeNode(3);
// assert true
System.out.println(Q101.instance.isSymmetric(root1));
TreeNode root2=new TreeNode(3);
root2.left=new TreeNode(4);
root2.right=new TreeNode(4);
root2.left.left=new TreeNode(5);
root2.right.right=new TreeNode(5);
root2.left.left.left=new TreeNode(6);
root2.right.right.right=new TreeNode(6);
// assert true
System.out.println(Q101.instance.isSymmetric(root2));
}
}
