[leetcode]101.Symmetric Tree

题目

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

解法一

思路

上来直接采用Same Tree的第二种解法，思路一模一样。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        Deque<TreeNode> queue = new LinkedList<>();
        queue.addLast(root.left);
        queue.addLast(root.right);
        while(!queue.isEmpty()){
            TreeNode l = queue.removeFirst();
            TreeNode r = queue.removeFirst();
            if(l == null && r == null) continue;
            else if(l == null || r == null || l.val != r.val)
                return false;
            else {
                queue.addLast(l.left);
                queue.addLast(r.right);
                queue.addLast(l.right);
                queue.addLast(r.left);
            }
        }
        return true;
    }
}

解法二

思路

递归的方法，一开始没有想出来，看了别人的解答发现构造一个辅助函数就容易理解很多。直接看代码就能理解思路。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isMirror(root.left, root.right);
    }
    
    public boolean isMirror(TreeNode p, TreeNode q) {
        if(p == null && q == null) return true;
        if(p == null || q == null || q.val != p.val)
            return false;
        return isMirror(p.left, q.right) && isMirror(p.right, q.left);
    }
}