101. Symmetric对称 Tree
Tree 对称 101
2023-09-11 14:14:21 时间
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
Note:
Bonus points if you could solve it both recursively and iteratively.
需要左右对称,解题思路和Same Tree类似。
如果拿掉root结点的话,逻辑其实相当于是比较root.left和root.right是否same tree类似的逻辑。只是需要把之前的left.left和right.right比较,left.right和right.left进行比较。
public bool IsSymmetric(TreeNode root) { return IsMirror(root?.left, root?.right); } public bool IsMirror(TreeNode left, TreeNode right) { bool flag; if (left == null && right == null) { flag = true; } else if (left == null || right == null) { flag = false; } else { if (left.val == right.val) { flag = IsMirror(left.left, right.right) && IsMirror(left.right, right.left); } else { flag = false; } } return flag; }
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