100. 相同的树 简单 == 遍历 递归和bfs

100. 相同的树
     给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。

如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

示例 1：

输入：p = [1,2,3], q = [1,2,3]
输出：true
示例 2：

输入：p = [1,2], q = [1,null,2]
输出：false
示例 3：

输入：p = [1,2,1], q = [1,1,2]
输出：false

提示：

两棵树上的节点数目都在范围 [0, 100] 内
-104 <= Node.val <= 104

递归

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

bfs

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
            Queue<TreeNode> qa=new LinkedList<>();
            Queue<TreeNode> qb=new LinkedList<>();
            qa.offer(p);
            qb.offer(q);
            while(!qa.isEmpty()&&!qb.isEmpty()){
                TreeNode a=qa.poll();
                TreeNode b=qb.poll();
                if(a==null&&b==null)continue;
                if(a==null||b==null)return false;
                if(a.val!=b.val)return false;
                qa.offer(a.left);qa.offer(a.right);
                qb.offer(b.left);qb.offer(b.right);
            }
            return qa.isEmpty()&&qb.isEmpty();
    }
}