Symmetric Tree

题目描述：

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

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

这个题用递归比较好处理，代码如下：

public boolean isSymmetric(TreeNode root) {
	if(root==null)
		return true;
	return BothIsSymmetric(root.left, root.right);
}

public boolean BothIsSymmetric(TreeNode root1,TreeNode root2) {
	if(root1==null&&root2==null)
		return true;
	if((root1==null^root2==null)||(root1!=null&&root2!=null&&root1.val!=root2.val))
		return false;
	return BothIsSymmetric(root1.left, root2.right)&&BothIsSymmetric(root1.right, root2.left);
}

关键是用迭代的方法去做

有的人说如果中序遍历之后对称那么原来的树就是镜像的，这简直就是误人子弟嘛。

因为[1,2,3,3,null,2,null]这种就不是镜像的！

然后看了其他人的做法，采用队列的方法去做。

但是我用的是ArrayDeque，这种offer，add，addLast方法（本质上都是调用addLast方法），都不能加入null值，导致程序出现各种各样的BUG，太浪费时间了。以后还是要多熟悉熟悉这些数据结构的源代码啊

enum Child{
	left,right;
}

class NewNode{
	TreeNode root;
	TreeNode parent;
	Child child;
	
	public NewNode(TreeNode root,TreeNode parent,Child child){
		this.root=root;
		this.parent=parent;
		this.child=child;
	}
}
public boolean isSymmetric(TreeNode root) {
	if(root==null)
		return true;
	Deque<NewNode> leftQueue=new ArrayDeque<NewNode>();
	Deque<NewNode> rightQueue=new ArrayDeque<NewNode>();
	if(root.left!=null)
		leftQueue.offer(new NewNode(root.left, root, Child.left));
	if(root.right!=null)
		rightQueue.offer(new NewNode(root.right, root, Child.right));
	while(!leftQueue.isEmpty()&&!rightQueue.isEmpty()){
		NewNode leftNewNode=leftQueue.poll();
		NewNode rightNewNode=rightQueue.poll();
		if(leftNewNode.root.val!=rightNewNode.root.val||leftNewNode.child==rightNewNode.child||leftNewNode.parent!=rightNewNode.parent)
			return false;
		if(leftNewNode.root.left!=null)
			leftQueue.offer(new NewNode(leftNewNode.root.left,leftNewNode.root,Child.left));
		if(leftNewNode.root.right!=null)
			leftQueue.offer(new NewNode(leftNewNode.root.right,leftNewNode.root,Child.right));
		if(rightNewNode.root.right!=null)
			rightQueue.offer(new NewNode(rightNewNode.root.right,rightNewNode.root,Child.right));
		if(rightNewNode.root.left!=null)
			rightQueue.offer(new NewNode(rightNewNode.root.left,rightNewNode.root,Child.left));
	}
	if(leftQueue.isEmpty()&&rightQueue.isEmpty())
		return true;
	return false;
}

于是我换成了LinkedList，这里的LinkedList使用的是双指针，实现了堆栈和队列，且可以加入null值。

正确迭代AC代码如下：

public boolean isSymmetric(TreeNode root) {
	if(root==null)
		return true;
	LinkedList<TreeNode> leftQueue=new LinkedList<TreeNode>();
	LinkedList<TreeNode> rightQueue=new LinkedList<TreeNode>();
	leftQueue.push(root.left);
	rightQueue.push(root.right);
	while(!leftQueue.isEmpty()&&!rightQueue.isEmpty()){
		TreeNode leftnode=leftQueue.poll();
		TreeNode rightnode=rightQueue.poll();
		if(leftnode==null&&rightnode==null)
			continue;
		if(leftnode==null||rightnode==null)
			return false;
		if(leftnode.val!=rightnode.val)
			return false;
		leftQueue.push(leftnode.left);
		leftQueue.push(leftnode.right);
		rightQueue.push(rightnode.right);
		rightQueue.push(rightnode.left);
	}
	if(leftQueue.isEmpty()&&rightQueue.isEmpty())
		return true;
	return false;
}