题目大意：判断一棵树是不是另外一棵树的子树。

https://leetcode.com/problems/subtree-of-another-tree/description/

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:

3
    / \
   4   5
  / \
 1   2

Given tree t:

4 
  / \
 1   2

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

Example 2:
Given tree s:

3
    / \
   4   5
  / \
 1   2
    /
   0

Given tree t:

4
  / \
 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);
  }
};

Related Problems

• [解题报告] Leetcode 100. Same Tree
• 花花酱 LeetCode 508. Most Frequent Subtree Sum
• 花花酱 LeetCode 563. Binary Tree Tilt

The post 花花酱 LeetCode 572. Subtree of Another Tree appeared first on Huahua's Tech Road.

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)

// 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); 
    }
};

// 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);
  }
}

"""
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

The post 花花酱 Leetcode 100. Same Tree appeared first on Huahua's Tech Road.