Leetcode——617. Merge Two Binary Trees

题目原址

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

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

给定两个二叉树t1,t2，要求将给定的两个二叉树对应节点的值相加组成一个新的二叉树，如果t1没有左孩子，t2有左孩子，则新二叉树对应节点的左孩子的值只为t2左孩子的值

解题思路

这是一道简单题，当然解题的思路也很简单，就是递归的经典思想，当前节点的值为t1和t2对应节点相加，当前节点的左右孩子分别递归调用该方法就可以了。

AC代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null && t2 == null)
            return null;
        else if(t1 == null)
            return t2;
        else if(t2 == null)
            return t1;
        TreeNode tn = new TreeNode(t1.val + t2.val);
        tn.left = mergeTrees(t1.left, t2.left);
        tn.right = mergeTrees(t1.right, t2.right);
        return tn;        
    }
}