Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1 / \ 2 2 \ \ 3 3
典型的因为是形状而递归(recursive)的题目。所以最好写个helper函数吧,更加方便表达。之后的参数可以事无巨细地传进去。这题的递归入口
就是左边、右边这样
这种可能是空,不一定有的节点,需要丢到递归函数中
//推测着写。 if ((root.left.left.val != root.right.right.val) || (root.left.right.val != root.right.left.val))