一开始想复杂了呀,没有做出来
直观的思路就是反转这颗树,然后判断两颗树是不是相同的
class Solution {
// 反转并创建一颗树
public TreeNode reverseTreeNode(TreeNode root) {
if (null == root) return null;
TreeNode tmp = new TreeNode(root.val);
tmp.left = reverseTreeNode(root.right);
tmp.right = reverseTreeNode(root.left);
return tmp;
}
public boolean isSame(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null) return true;
if (t1 == null || t2 == null) return false;
if (t1.val != t2.val) return false;
else return isSame(t1.left,t2.left) && isSame(t1.right,t2.right);
}
public boolean isSymmetric(TreeNode root) {
if (null == root) return true;
TreeNode newTree = reverseTreeNode(root);
return isSame(newTree,root);
}
}
网上大部分人的思路更为简洁,一开始没有想到。
即,判断左子树和右子树是否对称:
- 当前左右值是否相同
- 当前左子树的左子树和右子树的右子树是否相同
- 当前左子树的右子树和右子树的左子树是否相同
(画画图就看出来了)
利用了二叉树的特性把
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root == null) return true;
return isSymmetric(root.left, root.right);
}
public static boolean isSymmetric(TreeNode left, TreeNode right) {
if(left == null && right == null) return true;
if(left == null || right == null) return false;
if(left.val == right.val) return isSymmetric(left.left, right.right)
&& isSymmetric(left.right, right.left);
return false;
}
}