<leetcode>617. Merge Two Binary Trees

617. Merge Two Binary Trees

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.

 一开始的时候对题目的理解不清楚，不知道题目在说什么，就搁置了很久才写

今天又仔细看了一下题目，具体的意思就是在两个二叉树合并的过程中，相同层级的结点合并（值相加），对于一棵树对应层级如果一个有左子树，一个没有，就在最终的树种添加这个左子树；右子树同理。

遍历整个树时，中序，后序都无所谓的。

我用的是都添加到t1子树上，直接返回t1。

/** * 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;        if(t1==null&&t2!=null) return t2;        if(t1!=null&&t2==null) return t1;        if(t1!=null&&t2!=null){            t1.val = t1.val+t2.val;            t1.left = mergeTrees(t1.left,t2.left);            t1.right = mergeTrees(t1.right,t2.right);        }        return t1;    }}