[数据结构与算法]力扣100. 相同的树(javabfs)

?我的代码 很冗余

/**
 * 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==null && q!=null){
            return false;
        }
        Queue<TreeNode> quetree1 = new LinkedList<>();
        Queue<TreeNode> quetree2 = new LinkedList<>();
        quetree1.offer(p);
        quetree2.offer(q);
        while(!quetree1.isEmpty() && !quetree2.isEmpty()){
            TreeNode tree1 = quetree1.poll();
            TreeNode tree2 = quetree2.poll();
            if(tree1.val != tree2.val){
                return false;
            }
            if(tree1.left == null && tree2.left!=null){
                return false;
            }
            if(tree1.left != null && tree2.left==null){
                return false;
            }
            if(tree1.right == null && tree2.right!=null){
                return false;
            }
            if(tree1.right !=null && tree2.right ==null){
                return false;
            }
            if(tree1.left!=null && tree2.left!=null){
                quetree1.offer(tree1.left);
                quetree2.offer(tree2.left);
            }
            if(tree1.right!= null && tree2.right!=null){
                quetree1.offer(tree1.right);
                quetree2.offer(tree2.right);
            }
        }
        return quetree1.isEmpty()&&quetree2.isEmpty();
    }
}

答案代码

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        } else if (p == null || q == null) {
            return false;
        }
        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
        Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
        queue1.offer(p);
        queue2.offer(q);
        while (!queue1.isEmpty() && !queue2.isEmpty()) {
            TreeNode node1 = queue1.poll();
            TreeNode node2 = queue2.poll();
            if (node1.val != node2.val) {
                return false;
            }
            TreeNode left1 = node1.left, right1 = node1.right, left2 = node2.left, right2 = node2.right;
            if (left1 == null ^ left2 == null) {
                return false;
            }
            if (right1 == null ^ right2 == null) {
                return false;
            }
            if (left1 != null) {
                queue1.offer(left1);
            }
            if (right1 != null) {
                queue1.offer(right1);
            }
            if (left2 != null) {
                queue2.offer(left2);
            }
            if (right2 != null) {
                queue2.offer(right2);
            }
        }
        return queue1.isEmpty() && queue2.isEmpty();
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/same-tree/solution/xiang-tong-de-shu-by-leetcode-solution/
来源：力扣（LeetCode）
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^为异或。