LeetCode 617. Merge Two Binary Trees

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本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法。合并时，若节点重叠，则两个节点的值相加作为新节点的值；若节点不重叠，则采用非空节点作为新树的节点。示例展示了具体的合并过程。

Given two binary trees and 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 them 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 new tree.

Example 1:
Input: 
    Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
         3
        / \
       4   5
      / \   \ 
     5   4   7

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


    /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#define null NULL
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {

    TreeNode* p1=t1;
    TreeNode* p2=t2;
    TreeNode* p3=new TreeNode(0);
    if(p1!=null&&p2!=null){
        t1->val=t1->val+t2->val;
        t1->left=mergeTrees(t1->left,t2->left);
        t1->right=mergeTrees(t1->right,t2->right);
        return t1;
    }
    else if(p1==null&&p2!=null){
        p3->val=t2->val;
        t1=p3;
        t1->left=mergeTrees(t1->left,t2->left);
        t1->right=mergeTrees(t1->right,t2->right);
        return t1;
    }
    else if(p1!=null&&p1==null){   
        t1=t1;
        t1->left=mergeTrees(t1->left,t2->left);
        t1->right=mergeTrees(t1->right,t2->right);
        return t1;
    }
    else return t1;

    }
};