方法1:BFS
时间复杂度:$O(min(m, n))$
空间复杂度:$O(min(m, n))$
解题思路
根节点入队
循环步骤1-3直到队列为空
1. 节点出队
2. 节点的左子节点入队
3. 节点的右子节点入队
https://leetcode.cn/problems/same-tree/solution/xiang-tong-de-shu-by-leetcode-solution/
Java 代码
/**
* 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;
} else if (p == null || q == null) {
return false;
}
Queue<TreeNode> q1 = new LinkedList<>();
Queue<TreeNode> q2 = new LinkedList<>();
q1.offer(p);
q2.offer(q);
while (!q1.isEmpty() && !q2.isEmpty()) {
TreeNode node1 = q1.poll();
TreeNode node2 = q2.poll();
if (node1.val != node2.val) {
return false;
}
TreeNode left1 = node1.left, left2 = node2.left, right1 = node1.right, right2 = node2.right;
//如果结构不同直接false,不必再加入队列
if (left1 == null ^ left2 == null) {
return false;
}
if (right1 == null ^ right2 == null) {
return false;
}
if (left1 != null) {
q1.offer(left1);
}
if (right1 != null) {
q1.offer(right1);
}
if (left2 != null) {
q2.offer(left2);
}
if (right2 != null) {
q2.offer(right2);
}
}
return q1.isEmpty() && q2.isEmpty();
}
}
方法2:DFS
时间复杂度:$O(min(m, n))$
空间复杂度:$O(min(m, n))$
解题思路
递归,深度优先遍历。
https://leetcode.cn/problems/same-tree/solution/xiang-tong-de-shu-by-leetcode-solution/
Java 代码
/**
* 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;
} else if (p == null || q == null) {
return false;
} else if (p.val != q.val) {
return false;
} else {
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
}