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This approach utilizes recursion to traverse both trees simultaneously. If both node values are non-null, we sum them up as the new value and recursively merge their left and right children. If one node is null, we return the non-null node.
Time Complexity: O(n); Space Complexity: O(h), where n is the total number of nodes and h is the height of the tree.
1class TreeNode:
2 def __init__(self, x):
3 self.val = x
4 self.left = None
5 self.right = None
6
7def mergeTrees(t1, t2):
8 if not t1:
9 return t2
10 if not t2:
11 return t1
12 t1.val += t2.val
13 t1.left = mergeTrees(t1.left, t2.left)
14 t1.right = mergeTrees(t1.right, t2.right)
15 return t1This Python solution uses recursion to merge two binary trees. If either node is None, the function returns the other. Otherwise, it sums node values and recursively merges the child nodes.
This approach uses an iterative method with a stack to simulate the recursive calls for merging the two binary trees. We traverse the trees using a stack and merge nodes at each level iteratively.
Time Complexity: O(n); Space Complexity: O(h), where n is the total number of nodes and h is the maximum height of the two trees.
1#include <iostream>
#include <stack>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
if (!root1) return root2;
if (!root2) return root1;
stack<pair<TreeNode*, TreeNode*>> stk;
stk.push({root1, root2});
while (!stk.empty()) {
auto [n1, n2] = stk.top();
stk.pop();
if (!n1 || !n2) continue;
n1->val += n2->val;
if (!n1->left) {
n1->left = n2->left;
} else {
stk.push({n1->left, n2->left});
}
if (!n1->right) {
n1->right = n2->right;
} else {
stk.push({n1->right, n2->right});
}
}
return root1;
}This C++ solution uses a stack to iteratively merge nodes. We push pairs of nodes to the stack and at each step sum their values and push their left and right children if needed.