（Java）leetcode-101 Symmetric Tree

题目

判断对称树

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.

思路1

分治法：怎么判断一棵树是对称树？

• 该节点的左子树与右子树成镜像关系。

那么，如何判镜像是否成立？

• 两颗子树的根节点的值相等
• 一棵子树的左子树与另一子树的右子树成镜像关系
• 一棵子树的右子树与另一子树的左子树成镜像关系

分析到此很显然可以用分治法求解，时间复杂度O(n)，空间复杂度O(n)

代码（递归）

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root.left,root.right);
    }

    public boolean isMirror(TreeNode left,TreeNode right){
    	if(left == null || right == null)//退出条件，当其中一个是null时，若另一个也是null返回true，否则返回false
    		return left == right;
    	if(left.val != right.val)
    		return false;
    	return isMirror(left.left,right.right) && isMirror(left.right,right.left);
    }
}

提交结果

Runtime: 5 ms, faster than 100.00% of Java online submissions for Symmetric Tree.
Memory Usage: 39.5 MB, less than 16.77% of Java online submissions for Symmetric Tree.

思路2

采用层次遍历的方法，借助队列，每次出队需要比较的（处于对称位置上的）两个节点，判断值是否相等，如果相等，则继续将它们的左右子节点入队（这里的顺序要使对称位置的节点在队列中相邻），直到队列为空，说明树是对称的。
时间复杂度O(n)，空间复杂度O(n)

代码（迭代）

class Solution{
public boolean isSymmetric(TreeNode root) {
    if(root == null)
        return true;
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root.left);
    q.add(root.right);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}
}

提交结果

Runtime: 6 ms, faster than 72.49% of Java online submissions for Symmetric Tree.
Memory Usage: 41.4 MB, less than 5.12% of Java online submissions for Symmetric Tree.