首页 > 亚博体育登录 > [LeetCode]SymmetricTree

[LeetCode]SymmetricTree

2019-09-22

Problem

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

Example

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.

Solution

Recursion

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return dfs(root.left, root.right);
    }
    private boolean dfs(TreeNode left, TreeNode right) {
        if (left == null) return right == null;
        if (right == null) return left == null;
        if (left.val != right.val) {
            return false;
        }
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
}

Iteration

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Queue queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        while (!queue.isEmpty()) {
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if (left == null ^ right == null) return false;
            if (left == null && right == null) continue;
            if (left.val != right.val) return false;
            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);
        }
        return true;
    }
}

原文链接:https://segmentfault.com/a/1190000014663064

文章标签: