树3：请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的

对称树：

 //===================Mirror递归算法(理解递归)=============================//

通过求其镜像二叉树，然后判断这两树是否相同。几个子函数都用了递归。

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
import java.util.*;
public class Solution {
    void Mirror(TreeNode root){//求树的镜像树
        if(root==null) return;
        Mirror(root.left);
        Mirror(root.right);
        if(root.left!=null||root.right!=null)
        {
            TreeNode temp=root.left;
            root.left=root.right;
            root.right=temp;
        }
    }
    boolean isSameTree(TreeNode t1,TreeNode t2){//判断两棵树是否相同
        if(t1==null&&t2==null) return true;
        else if(t1!=null&&t2!=null&&t1.val==t2.val){
            boolean left=isSameTree(t1.left,t2.left);
            boolean right=isSameTree(t1.right,t2.right);
            return left&&right;
        }else
            return false;
    }
    TreeNode copyTree(TreeNode root){//复制一棵树
        if(root==null) return null;
        TreeNode t=new TreeNode(root.val);
        t.left=copyTree(root.left);
        t.right=copyTree(root.right);
        return t;
    }
    boolean isSymmetrical(TreeNode pRoot)
    {
        TreeNode temp=copyTree(pRoot);
        Mirror(pRoot);
        return isSameTree(temp,pRoot);
    }
}

//===================isMirror递归算法2=============================//

1.只要pRoot.left和pRoot.right对称即可

2.左右节点的值相等且对称子树left.left， right.right ;left.rigth,right.left也对称

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public boolean jude(TreeNode node1,TreeNode node2){
        if(node1==null&&node2==null){
            return true;
        }else if(node1==null||node2==null)
            return false;
        if(node1.val!=node2.val)
            return false;
        else{
            return jude(node1.left,node2.right)&&jude(node1.right,node2.left);
        }
    }
    public boolean isSymmetrical(TreeNode pRoot)
    {
        return pRoot==null||jude(pRoot.left,pRoot.right);
    }
}

//===================非递归算法，利用DFS和BFS=============================//

 DFS:成对出栈入栈

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
import java.util.*;
//队列的进队出队：Q.offer(object) Q.poll()
public class Solution {
    boolean isSymmetrical(TreeNode pRoot)
    {
        if(pRoot == null) return true;//1.树null时直接返回true
        //2.树不空
        Stack<TreeNode> s = new Stack<>();
        s.push(pRoot.left);
        s.push(pRoot.right);
        while(!s.empty()) {
            TreeNode right = s.pop();//成对取出
            TreeNode left = s.pop();
            //首先是要结构对称
            if(left == null && right == null) continue;
            else if(left == null || right == null) return false;
            //然后要保证对称位置上的数据相等
            if(left.val != right.val) return false;
            else{
                s.push(left.left);//成对插入 left.left对应right.right ;left.rigth对应right.left
                s.push(right.right);
                s.push(left.right);
                s.push(right.left);
            }
        }
        return true;
    }
}

 BFS：

import java.util.*;
//队列的进队出队：Q.offer(object) Q.poll()
public class Solution {
    boolean isSymmetrical(TreeNode pRoot)
    {
        if(pRoot==null)
            return true;
        Queue<TreeNode> Q=new LinkedList<>();
        Q.offer(pRoot.left);//成对入队
        Q.offer(pRoot.right);
        while(!Q.isEmpty()){
            TreeNode left=Q.poll();//成对出队
            TreeNode right=Q.poll();
            //首先是要结构对称
            if(left==null&&right==null)
                continue;
            else if(left==null||right==null)
                return false;
            //然后要保证对称位置上的数据相等
            if(left.val!=right.val) return false;
            else{
                Q.offer(left.left);
                Q.offer(right.right);
                Q.offer(left.right);
                Q.offer(right.left);
            }
        }
        return true;
    }
}