Problem

题目大意：判断一棵树是不是另外一棵树的子树。

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

/ \

4 5

/ \

1 2

Given tree t:

/ \

1 2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

/ \

4 5

/ \

1 2

Given tree t:

/ \

1 2

Return false.

Solution: Recursion

Time complexity: O(max(n, m))

Space complexity: O(max(n, m))

C++

// Author: Huahua

// Running time: 29 ms

class Solution {

public:

bool isSubtree(TreeNode* s, TreeNode* t) {

if (t == nullptr) return true;

if (s == nullptr) return false;

if (isSameTree(s, t)) return true;

return isSubtree(s->left, t) || isSubtree(s->right, t);

}

private:

bool isSameTree(TreeNode* s, TreeNode* t) {

if (s == nullptr && t == nullptr) return true;

if (s == nullptr || t == nullptr) return false;

return (s->val == t->val)

&& isSameTree(s->left, t->left)

&& isSameTree(s->right, t->right);

}

};

Problem:

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

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

bool isSameTree(TreeNode* p, TreeNode* q) {

// Both are emtry: same

if (!p && !q) return true;

// One is emtry: not the Same

if (!p || !q) return false;

// Both are not emptry, compare the root value

if (p->val != q->val) return false;

// Compare left-subtree and right-subtree recursively.

return isSameTree(p->left, q->left)

&& isSameTree(p->right, q->right);

}

};

Java

// Author: Huahua

class Solution {

public boolean isSameTree(TreeNode p, TreeNode q) {

if (p == null && q == null) return true;

if (p == null || q == null) return false;

return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);

}

Python3

"""

Author: Huahua

"""

class Solution:

def isSameTree(self, p, q):

if not p and not q: return True

if not p or not q: return False

return all((p.val == q.val,

self.isSameTree(p.left, q.left),

self.isSameTree(p.right, q.right)))

Follow Up

花花酱 LeetCode 572. Subtree of Another Tree

Posts tagged as “same tree”

花花酱 LeetCode 572. Subtree of Another Tree

Problem

Solution: Recursion

Related Problems

花花酱 Leetcode 100. Same Tree

Problem:

Example 1:

Example 2:

Example 3:

Solution: Recursion

C++

Java

Python3

Follow Up