LeetCode Merge Two Trees-优快云博客

本文链接：https://blog.youkuaiyun.com/sun_wangdong/article/details/77487711

本文介绍了一种将两棵二叉树合并的算法，当两棵树的节点重叠时，节点值相加；否则保留非空节点。通过递归方式实现了这一过程，并给出了具体的代码示例。

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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

题解：

此题是将两棵二叉树进行合并，当遇到相同位置的节点时，将两棵树的相同位置节点的数值进行求和，针对二叉树的题目，一般情况下，我们采用递归的方式来做，针对此题，可以先把t1树保持不变，然后从t2树的根节点开始，将两棵树的根节点的值相加，然后分别针对根节点的左子树和右子树，递归调用函数，可以实现将两棵树相加的功能。具体代码如下所示：

class TreeNode
{
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x)
    {
        val = x;
    }
}

public class MergeTrees
{
    public TreeNode mergeTrees(TreeNode t1,TreeNode t2)
    {
        if(t1 == null && t2 == null)   //这里分三种情况来讨论，两棵树的空与否情况
            return null;
        else if(t1 == null && t2 != null)
            return t2;
        else if(t1 != null && t2 == null)
            return t1;
        else
        {
            t1.val += t2.val;
            t1.left = mergeTrees(t1.left,t2.left);                         //左子树递归调用
            t1.right = mergeTrees(t1.right,t2.right);                      //右子树递归调用
            return t1;
        }
        //return t1;
    }
}