leetcode617

这道题想了很久，并没有掌握思想，写了很多，也没有解决。先贴出思考的过程。

class Solution {
public:
    vector<TreeNode> v1;
    vector<TreeNode> v2;

    queue<TreeNode> Q1;
    queue<TreeNode> Q2;

    void FloorTree(TreeNode tree, int type)
    {
        TreeNode node = TreeNode(tree.val);
        node.left = tree.left;
        node.right = tree.right;
        if (type == 1)
        {
            Q1.push(node);
            while (!Q1.empty())
            {
                TreeNode livenode = TreeNode(0);
                livenode = Q1.front();
                Q1.pop();
                v1.push_back(livenode);

                if (&livenode != NULL && (livenode.left != NULL || livenode.right != NULL))
                {
                    if (livenode.left != NULL)
                    {
                        Q1.push(*livenode.left);
                    }
                    else
                    {
                        TreeNode node = TreeNode(0);
                        Q1.push(node);
                    }

                    if (livenode.right != NULL)
                    {
                        Q1.push(*livenode.right);
                    }
                    else
                    {
                        TreeNode node = TreeNode(0);
                        Q1.push(node);
                    }
                }
            }
        }
        else
        {
            Q2.push(node);
            while (!Q2.empty())
            {
                TreeNode livenode = TreeNode(0);
                livenode = Q2.front();
                Q2.pop();
                v2.push_back(livenode);
                if (&livenode != NULL && (livenode.left != NULL || livenode.right != NULL))
                {
                    if (livenode.left != NULL)
                    {
                        Q2.push(*livenode.left);
                    }
                    else
                    {
                        TreeNode node = TreeNode(0);
                        Q2.push(node);
                    }

                    if (livenode.right != NULL)
                    {
                        Q2.push(*livenode.right);
                    }
                    else
                    {
                        TreeNode node = TreeNode(0);
                        Q2.push(node);
                    }
                }
            }
        }
    }

    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        FloorTree(*t1, 1);
        FloorTree(*t2, 2);

        int len1 = v1.size();
        int len2 = v2.size();

        vector<TreeNode> V;
        if (len1 <= len2)
        {
            for (int i = 0; i < len1; i++)
            {
                TreeNode t1 = v1[i];
                TreeNode t2 = v2[i];
                t1.val += t2.val;
                V.push_back(t1);
            }
            for (int i = len1; i < len2; i++)
            {
                TreeNode t = v2[i];
                V.push_back(t);
            }
        }
        else
        {
            for (int i = 0; i < len2; i++)
            {
                TreeNode t1 = v1[i];
                TreeNode t2 = v2[i];
                t2.val += t1.val;
                V.push_back(t2);
            }
            for (int i = len2; i < len1; i++)
            {
                TreeNode t = v1[i];
                V.push_back(t);
            }
        }

        for (int i = 0; i < V.size(); i++)
        {
            
        }

        TreeNode* N = NULL;
        return N;
    }
};

下面列出正确的解法。

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (t1 == NULL)return t2;
        if (t2 == NULL)return t1;
        t1->val += t2->val;
        t1->left = mergeTrees(t1->left, t2->left);
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }

补充一个python的实现，和上面的C++的思路基本一样，只是我个人感觉更容易理解一些：

 1 class Solution:
 2     def mergeTrees(self, t1: 'TreeNode', t2: 'TreeNode') -> 'TreeNode':
 3         if t1!=None or t2!=None:
 4             if t1==None:
 5                 t1 = TreeNode(0)
 6             if t2==None:
 7                 t2 = TreeNode(0)
 8             t = TreeNode(0)
 9             t.val = t1.val + t2.val
10             t.left = self.mergeTrees(t1.left,t2.left)
11             t.right = self.mergeTrees(t1.right,t2.right)
12             return t
13 
14         return None

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原文地址：https://www.cnblogs.com/asenyang/p/9706343.html