likou617合并二叉树

本体正宗未看题解！本人自己写出来的！！！嘿嘿嘿

思路很简单，传入两个二叉树根节点，分别有三个细节判断：

第一：a二叉树为空，b二叉树不为空时，继续递归b二叉树

第二：a二叉树不为空，b二叉树为空，继续递归a二叉树

第三：全部为空，返回空节点

还有一个不算细节，是递归的必须条件，递归循环体

package shuJuJieGouYuSuanFa.erChaShu;
 
 
import java.util.LinkedList;
import java.util.Queue;
 
public class likou617 {
    public static void main(String[] args) {
        TreeNode r1 = new TreeNode(1);
        TreeNode r2 = new TreeNode(1);
        r1.left = new TreeNode(2);
        r1.left.left = new TreeNode(3);
        r2.right = new TreeNode(2);
        r2.right.right = new TreeNode(3);
        TreeNode t = new likou617().mergeTrees(r1, r2);
//        f1(t);
        f2(t);
    }
 
    private static void f2(TreeNode t) {
        Queue dui = new LinkedList<>();
        int size = 0;
        TreeNode p = null;
        dui.offer(t);
        while (!dui.isEmpty()) {
            size = dui.size();
            for (int i = 0; i < size; i++) {
                p = dui.poll();
                System.out.print(p.val + " ");
                if (p.left != null) {
                    dui.offer(p.left);
                }
                if (p.right != null) {
                    dui.offer(p.right);
                }
            }
            System.out.println();
        }
    }
 
    private static void f1(TreeNode t) {
        if (t == null) {
            System.out.print("null ");
            return;
        }
        System.out.print(t.val + " ");
        f1(t.left);
        f1(t.right);
    }
 
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        return dfs(root1, root2);
 
    }
 
    private TreeNode dfs(TreeNode root1, TreeNode root2) {
        TreeNode p = null;
        if (root1 == null && root2 != null) {
            p = new TreeNode(root2.val);
            p.left = dfs(root1, root2.left);
            p.right = dfs(root1, root2.right);
            return p;
        } else if (root1 != null && root2 == null) {
            p = new TreeNode(root1.val);
            p.left = dfs(root1.left, root2);
            p.right = dfs(root1.right, root2);
            return p;
        } else if (root1 == null && root2 == null) {
            return null;
        }
        p = new TreeNode(root1.val + root2.val);
        p.left = dfs(root1.left, root2.left);
        p.right = dfs(root1.right, root2.right);
        return p;
    }
}

 自己写出来还是挺有成就感的，嘿嘿嘿