LeetCode 100. Same Tree(相同树)
题目描述
Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
Example 1:
Input: 1 1
/ \ / \
2 3 2 3
[1,2,3], [1,2,3]
Output: true
Example 2:
Input: 1 1
/ \
2 2
[1,2], [1,null,2]
Output: false
Example 3:
Input: 1 1
/ \ / \
2 1 1 2
[1,2,1], [1,1,2]
Output: false
解法一
思路
既然是树的遍历问题,那又是递归了,只不过这里要同步递归两颗树。递归的时候,对于当前的两颗树上的结点,首先比较其结点是否相同,即是否同时为null
或值相等;其次如果值相等,则递归比较其左右子树。
换句话说,同步递归两颗树保证了两颗树上当前结点位置的一致性,而对两颗树当前结点的比较,则保证了当前结点的一致性。
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
if (p is None and q is None) \
or (p is not None and q is not None and p.val == q.val
and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)):
return True
return False
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if ((p == null && q == null) ||
(p != null && q != null && p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right))) {
return true;
}
return false;
}
}
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSameTree(TreeNode *p, TreeNode *q) {
if ((p == nullptr && q == nullptr) ||
(p != nullptr && q != nullptr && p->val == q->val && isSameTree(p->left, q->left) &&
isSameTree(p->right, q->right))) {
return true;
}
return false;
}
};