题解 | #合并二叉树#TOP32

思路:
1.递归遍历
2.递归终止条件，只要有任意节点递归完了叶子结点。每个节点等于当前两个节点的值，否则就直接指向其中一个节点

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */
    public TreeNode mergeTrees (TreeNode t1, TreeNode t2) {
        // write code here
        if(t1 == null && t2 == null){
            return null;
        }
        if(t1 == null){
            return t2;
        }
        if(t2 == null){
            return t1;
        }
        t1.val = t1.val + t2.val;
        t1.left = mergeTrees(t1.left, t2.left);
        t1.right = mergeTrees(t1.right, t2.right);
        return t1;
    }
    
}