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Symmetric Tree

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.

public class Symmetric {
    public static void main(String[] args) {

    }

    public boolean isSymmetric(TreeNode root) {

        if (root == null) return true;

        return checker(root.left, root.right);
    }

    public boolean checker(TreeNode left, TreeNode right) {
        if (left == null && right != null || left != null && right == null) return false;
        if (left == null && right == null) return true;

        if (left.val != right.val) return false;

        return checker(left.left, right.right) && checker(left.right, right.left);
    }
}

/*
执行用时 :0 ms, 在所有 Java 提交中击败了100.00%的用户
内存消耗 :37.4 MB, 在所有 Java 提交中击败了41.58%的用户
*/

The base case is that if (root == null).
First divide the problem into smallest piece that we judge the node’s value equals or not.
Then we conquer the problem that the node left.left equals right.right and left.right equals right.left.

Reference
https://leetcode-cn.com/problems/symmetric-tree/submissions