LeetCode-617. Merge Two Binary Trees [C++][Java]

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最新推荐文章于 2025-08-17 23:28:34 发布

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分类专栏： LeetCode刷题怪文章标签： leetcode c++ 算法

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本文介绍如何解决LeetCode上的合并二叉树问题，即给定两个二叉树，需要将它们合并成一棵新的二叉树。如果两棵树中存在相同的节点，则该节点的值为两棵树上该节点的值之和；如果只有一棵树中有该节点，则该节点会出现在合并后的新树中。

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LeetCode-617. Merge Two Binary TreesLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/merge-two-binary-trees/

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].
-10^4 <= Node.val <= 10^4

【C++】

/**
 * 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) {
        TreeNode* root = new TreeNode();
        if (root1 && root2) {
            root->val = root1->val + root2->val;
            root->left = mergeTrees(root1->left, root2->left);
            root->right = mergeTrees(root1->right, root2->right);
        } else if (root1) {
            root->val = root1->val;
            root->left = mergeTrees(root1->left, nullptr);
            root->right = mergeTrees(root1->right, nullptr);
        } else if (root2) {
            root->val = root2->val;
            root->left = mergeTrees(nullptr, root2->left);
            root->right = mergeTrees(nullptr, root2->right);
        } else {
            return nullptr;
        }
        return root;
    }
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        TreeNode root = new TreeNode();
        if (root1 != null && root2 != null) {
            root.val = root1.val + root2.val;
            root.left = mergeTrees(root1.left, root2.left);
            root.right = mergeTrees(root1.right, root2.right);
        } else if (root1 != null) {
            root.val = root1.val;
            root.left = mergeTrees(root1.left, null);
            root.right = mergeTrees(root1.right, null);
        } else if (root2 != null) {
            root.val = root2.val;
            root.left = mergeTrees(null, root2.left);
            root.right = mergeTrees(null, root2.right);
        } else {
            return null;
        }
        return root;
    }
}