Merge Two Binary Trees[E,TreeDepth-First Search,Breadth-First Search,Binary Tree]

https://leetcode.com/problems/merge-two-binary-trees/description/

You are given two binary trees root1 and root2.

Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.

Return the merged tree.

Note: The merging process must start from the root nodes of both trees.

 

Example 1:

Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
Output: [3,4,5,5,4,null,7]

Example 2:

Input: root1 = [1], root2 = [1,2]
Output: [2,2]

 

Constraints:

• The number of nodes in both trees is in the range [0, 2000].
• -104 <= Node.val <= 104

처음 보는 신박한 Tree Merge 문제다.

 

1. C++

어렵다..아

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if(!root1 && !root2) return nullptr;

        int val = (root1 ? root1->val : 0) + (root2 ? root2->val : 0);
        TreeNode* ans = new TreeNode(val);

        ans->left = mergeTrees(root1 ? root1->left:nullptr,root2 ? root2->left:nullptr);
        ans->right = mergeTrees(root1 ? root1->right:nullptr,root2 ? root2->right:nullptr);

        return ans;
    }
};

 

2. C 유사하게 풀었다.

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* mergeTrees(struct TreeNode* root1, struct TreeNode* root2) {
    if(!root1) return root2;
    else if(!root2) return root1;

    struct TreeNode* ans = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    ans->val = root1->val + root2->val;

    ans->left = mergeTrees(root1->left,root2->left);
    ans->right = mergeTrees(root1->right,root2->right);

    return ans;
}

 

3. Python #BFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root1 : return root2
        elif not root2 : return root1

        q = deque()
        q.append((root1,root2))

        while q:
            curr1,curr2 = q.popleft()
            curr1.val += curr2.val

            if curr1.left and curr2.left :
                q.append((curr1.left,curr2.left))
            elif not curr1.left:
                curr1.left = curr2.left
            
            if curr1.right and curr2.right :
                q.append((curr1.right,curr2.right))
            elif not curr1.right:
                curr1.right = curr2.right
        
        return root1