Leetcode代码学习周记——Merge Two Binary Trees

最新推荐文章于 2018-10-29 19:30:22 发布

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本文介绍了一种算法，用于解决LeetCode上的一道题目——合并两个二叉树。通过递归的方式，对两个输入的二叉树进行节点求和合并，并详细展示了具体的实现过程。

题目链接：

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

题目描述：

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

对给定两个二叉树进行求和合并。

题解：

对同一位置的结点：1.若两棵树该位置节点均为空，则返回空 2.若有一棵树该节点为空，则返回另一棵树 3.否则,将两个结点的值相加后的结点返回。

然后分别对1和2的左右子树进行递归调用。

代码：

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (t1 == NULL && t2 == NULL) return NULL;
        if (t1 == NULL) return t2;
        if (t2 == NULL) return t1;
        TreeNode *t = new TreeNode(t1->val + t2->val);
        t->left = mergeTrees(t1->left, t2->left);
        t->right = mergeTrees(t1->right, t2->right);
        return t;
    }
};