Leetcode - Symmetric Tree

Question:

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

《Leetcode - Symmetric Tree》 Paste_Image.png

But the following is not:

《Leetcode - Symmetric Tree》 Paste_Image.png
《Leetcode - Symmetric Tree》 Paste_Image.png

My code:

/**
 * 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 isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        else
            return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null && right == null)
            return true;
        else if (left != null && right == null)
            return false;
        else if (left == null && right != null)
            return false;
        else {
            return (left.val == right.val) && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);

        }
    }
}

My test result:

《Leetcode - Symmetric Tree》 Paste_Image.png

**
总结:
这次作业也不是很难吧。
所以一个小时做完了两道 easy级别的题目。
真没啥总结的,就是递归吧。写递归的能力被之前上算法课培训出来了,比较简单的都还是能写写的。。。
加油。

**

Good luck, Richardo!

My code:

/**
 * 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 isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            ArrayList<TreeNode> l = new ArrayList<TreeNode>(size);
            while (!q.isEmpty())
                l.add(q.poll());
            for (int i = 0; i < size / 2; i++) {
                TreeNode left = l.get(i);
                TreeNode right = l.get(size - i - 1);
                if (left.val != right.val)
                    return false;
                if (left.left == null && right.right != null)
                    return false;
                if (left.left != null && right.right == null)
                    return false;
                if (left.right == null && right.left != null)
                    return false;
                if (left.right != null && right.left == null)
                    return false;
            }
            for (int i = 0; i < size; i++) {
                TreeNode temp = l.get(i);
                if (temp.left != null)
                    q.offer(temp.left);
                if (temp.right != null)
                    q.offer(temp.right);
            }
            if (q.size() % 2 == 1)
                return false;
        }
        return true;
    }
}

用层序遍历实现了非递归解法。但是时间好长。
用一个队列,存放已经被检查过的结点,按从左往右的顺序。
然后每次进循环,把他们输入进一个list,然后从两端往中间,一个结点一个结点得扫描。判断他们的下一层在结构上是否满足对称。还要检查他们自己的value是否相等。
如果都满足条件。按序把它们的下一层输入进队列。
然后就差不多了。

递归做法:
My code:

/**
 * 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 isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        else if (root.left == null && root.right == null)
            return true;
        else if (root.left == null || root.right == null)
            return false;
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left.val != right.val)
            return false;
        boolean b1 = false;
        boolean b2 = false;
        if (left.left == null && right.right == null)
            b1 = true;
        else if (left.left == null || right.right == null)
            return false;
        else
            b1 = isSymmetric(left.left, right.right);
        if (!b1)
            return false;
        
        if (left.right == null && right.left == null)
            b2 = true;
        else if (left.right == null || right.left == null)
            return false;
        else 
            b2 = isSymmetric(left.right, right.left);
        
        return b2;
    }
}

意思差不多。

昨晚打电话到两点。又早起上课,一直没停。所以晚上的时候特别累。还得跟着去上课。所以人比较暴躁,心态不好。
晚上回去好好休息。
又拿到了一个OA。好好珍惜。
最近开始练习 Tree。

Anyway, Good luck, Richardo!

My code:

/**
 * 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 isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }
    
    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        else if (left == null || right == null) {
            return false;
        }
        else if (left.val != right.val) {
            return false;
        }
        else {
            return helper(left.left, right.right) && helper(left.right, right.left);
        }
    }
}

recursion 不是很难。
看了上面的 iteration解法,感觉有点烦。
然后看了下答案:
My code:

/**
 * 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 isSymmetric(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        q.add(root);
        while (!q.isEmpty()) {
            TreeNode n1 = q.poll();
            TreeNode n2 = q.poll();
            if (n1 == null && n2 == null) {
                continue;
            }
            else if (n1 == null || n2 == null) {
                return false;
            }
            else if (n1.val != n2.val) {
                return false;
            }
            else {
                q.offer(n1.left);
                q.offer(n2.right);
                q.offer(n1.right);
                q.offer(n2.left);
            }
        }
        
        return true;
    }
}

reference:
https://leetcode.com/articles/symmetric-tree/

这种层序遍历和之前的不太一样。他是每次压入两个,弹出两个。
Interesting

当分析 recursion 复杂度时,
分析空间的时候,记住考虑到 recursive call 需要在栈中占的体积。
比如这道题目,在最恶劣的情况下,如果树是一条线,那么它所占的体积就是 O(n)

Anyway, Good luck, Richardo! — 08/28/2016

    原文作者:Richardo92
    原文地址: https://www.jianshu.com/p/6f1dcbbb4602#comments
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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