本系列导航：剑指offer（第二版）java实现导航帖

面试题28：对称的二叉树

题目要求：
判断一棵二叉树是不是对称的。如果某二叉树与它的镜像一样，称它是对称的。

解题思路：
比较直接的思路是比较原树与它的镜像是否一样。书中就是用的这种方式（比较二叉树的前序遍历和对称前序遍历）。但这种思路下，树的每个节点都要读两次，也就是遍历两遍。
其实可以只遍历一次完成判断：我们可以通过判断待判断二叉树的左子树与右子树是不是对称的来得知该二叉树是否是对称的。

package structure;
import java.util.LinkedList;
import java.util.Queue;
/**
 * Created by ryder on 2017/6/12.
 * 树节点
 */
public class TreeNode<T> {
    public T val;
    public TreeNode<T> left;
    public TreeNode<T> right;
    public TreeNode(T val){
        this.val = val;
        this.left = null;
        this.right = null;
    }
    //层序遍历
    @Override
    public String toString() {
        StringBuilder stringBuilder = new StringBuilder("[");
        Queue<TreeNode<T>> queue = new LinkedList<>();
        queue.offer(this);
        TreeNode<T> temp;
        while(!queue.isEmpty()){
            temp = queue.poll();
            stringBuilder.append(temp.val);
            stringBuilder.append(",");
            if(temp.left!=null)
                queue.offer(temp.left);
            if(temp.right!=null)
                queue.offer(temp.right);
        }
        stringBuilder.deleteCharAt(stringBuilder.lastIndexOf(","));
        stringBuilder.append("]");
        return stringBuilder.toString();
    }
}

package chapter4;
import structure.TreeNode;
import java.util.LinkedList;
import java.util.Queue;
/**
 * Created by ryder on 2017/7/15.
 * 对称的二叉树
 */
public class P159_SymmetricalBinaryTree {
    //递归实现
    public static boolean isSymmetrical(TreeNode<Integer> root){
        if(root==null)
            return true;
        if(root.left==null && root.right==null)
            return true;
        if(root.left==null || root.right==null)
            return false;
        return isSymmetrical(root.left,root.right);
    }
    public static boolean isSymmetrical(TreeNode<Integer> root1,TreeNode<Integer> root2){
        if(root1==null && root2==null)
            return true;
        if(root1==null || root2==null)
            return false;
        if(!root1.val.equals(root2.val))
            return false;
        return isSymmetrical(root1.left,root2.right) && isSymmetrical(root1.right,root2.left);
    }
    //迭代实现
    public static boolean isSymmetrical2(TreeNode<Integer> root){
        if(root==null)
            return true;
        if(root.left==null && root.right==null)
            return true;
        if(root.left==null || root.right==null)
            return false;
        Queue<TreeNode<Integer>> queueLeft = new LinkedList<>();
        Queue<TreeNode<Integer>> queueRight = new LinkedList<>();
        queueLeft.offer(root.left);
        queueRight.offer(root.right);
        TreeNode<Integer> tempLeft,tempRight;
        while(!queueLeft.isEmpty()|| !queueRight.isEmpty()){
            tempLeft = queueLeft.poll();
            tempRight = queueRight.poll();
            if(tempLeft.val.equals(tempRight.val)){
                if(tempLeft.left!=null)
                    queueLeft.offer(tempLeft.left);
                if(tempLeft.right!=null)
                    queueLeft.offer(tempLeft.right);
                if(tempRight.right!=null)
                    queueRight.offer(tempRight.right);
                if(tempRight.left!=null)
                    queueRight.offer(tempRight.left);
            }
            else
                return false;
        }
        if(queueLeft.isEmpty() && queueRight.isEmpty())
            return true;
        else
            return false;
    }
    public static void main(String[] args){
        TreeNode<Integer> root = new TreeNode<>(8);
        root.left = new TreeNode<>(6);
        root.right = new TreeNode<>(6);
        root.left.left = new TreeNode<>(5);
        root.left.right = new TreeNode<>(7);
        root.right.left = new TreeNode<>(7);
        root.right.right = new TreeNode<>(5);
        System.out.println(isSymmetrical(root));
        System.out.println(isSymmetrical2(root));
    }
}

运行结果

true
true