【题解】【LeetCode】100. 相同的树

原题传送：链接

100. 相同的树

给定两个二叉树，编写一个函数来检验它们是否相同。
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:       1         1
          / \       / \
         2   3     2   3

		[1,2,3],   [1,2,3]
		
输出: true

示例 2:

输入:      1          1
          /           \
         2             2

        [1,2],     [1,null,2]

输出: false

示例 3:

输入:       1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

输出: false

思路：

递归遍历，若当前两个结点都为空，则返回真；若两个结点一空一非空，则返回假；若两个结点都为非空，则比较数值是否相等。

C:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

bool isSameTree(struct TreeNode* p, struct TreeNode* q){
    if(!p && !q)
        return true;
    if(!q || !p)
        return false;
    if(p->val != q->val)
        return false;
    return isSameTree(p->right, q->right) && isSameTree(p->left, q->left);
}

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 && !q)
            return true;
        if(!q || !p)
            return false;
        if(p->val != q->val)
            return false;
        return isSameTree(p->right, q->right) && isSameTree(p->left, q->left);
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
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.right, q.right) && isSameTree(p.left, q.left);
    }
}

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
        if not p and not q:
            return True
        if not q or not p:
            return False
        if p.val != q.val:
            return False
        return self.isSameTree(p.right, q.right) and self.isSameTree(p.left, q.left)