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Leetcode Same Tree problem solution

YASH PAL, 31 July 2024

In this Leetcode Same Tree problem solution we have Given the roots of two binary trees p and q, 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.

Leetcode Same Tree problem solution

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  • Problem solution in Python.
  • Problem solution in Java.
  • Problem solution in C++.
  • Problem solution in C.

Problem solution in Python.

from itertools import zip_longest

class Solution:
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        for p_val, q_val in zip_longest(preorder(p), preorder(q)):
            if p_val != q_val:
                return False
        return True
    
def preorder(tree):
        yield tree and tree.val
        if tree:
            for node in preorder(tree.left):
                yield node
            for node in preorder(tree.right):
                yield node

Problem solution in Java.

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p != null && q != null && q.val == p.val) {
            return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
        }
        return p == null && q == null;
    }
}

Problem solution in C++.

class Solution {
public:
    bool 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);
    }
};

Problem solution in C.

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

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