思路:递归判断。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p==null&&q==null) {
return true;
}else if (p==null||q==null) {
return false;
}
if (p.val!=q.val) {
return false;
}else{
return isSameTree(p.left, q.left)&&isSameTree(p.right, q.right);
}
}
}
耗时:260ms,另外发现if不写中括号的写法比写中括号的写法来的快。