LeetCode ：617. 合并二叉树（java版）

力扣

思路不难，下笔难...

我泰菜了(灬ꈍ ꈍ灬)

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        // 1.广度优先搜索
        if(root1==null) return root2;
        if(root2==null) return root1;
        Queue queue=new LinkedList<>();
        Queue queue1=new LinkedList<>();
        Queue queue2=new LinkedList<>();
        TreeNode merge=new TreeNode(root1.val+root2.val);
        queue.offer(merge);
        queue1.offer(root1);
        queue2.offer(root2);
        while(!queue.isEmpty()){
            TreeNode node=queue.poll(), node1=queue1.poll(), node2=queue2.poll();
            TreeNode left1=node1.left, right1=node1.right, left2=node2.left, right2=node2.right;
            if(left1!=null && left2!=null){
                TreeNode tmp=new TreeNode(left1.val+left2.val);
                node.left=tmp;
                queue.offer(tmp);
                queue1.offer(left1);
                queue2.offer(left2);
            }
            if(left1==null)  node.left=left2;
            if(left2==null)  node.left=left1;
            if(right1!=null && right2!=null){
                TreeNode tmp=new TreeNode(right1.val+right2.val);
                node.right=tmp;
                queue.offer(tmp);
                queue1.offer(right1);
                queue2.offer(right2);
            }
            if(right1==null)    node.right=right2;
            if(right2==null)    node.right=right1;
                
            
        }
        return merge;

    }
}

不创建新的树，直接在root1上作变动(挂载)：

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        // 3.广度优先——不创建新的树，直接在root1上作变动(挂载)
        if(root1==null || root2==null)
            return root1==null?root2:root1;
        Queue queue=new LinkedList<>();
        queue.offer(root1);
        queue.offer(root2);
        while(!queue.isEmpty()){
            TreeNode node1=queue.poll();
            TreeNode node2=queue.poll();
            node1.val+=node2.val;
            if(node1.left!=null && node2.left!=null){
                queue.offer(node1.left);
                queue.offer(node2.left);
            }
            if(node1.left==null)    node1.left=node2.left;
            if(node1.right!=null && node2.right!=null){
                queue.offer(node1.right);
                queue.offer(node2.right);
            }
            if(node1.right==null)   node1.right=node2.right;
            
        }
        return root1;

    }
}

package linkedList;

import java.time.LocalDate;
import java.util.*;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}
class Solution {
//    Definition for a binary tree node.


    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        // 2.深度优先搜索
        if(root1==null) return root2;
        if(root2==null) return root1;
        TreeNode merge=new TreeNode(root1.val+root2.val);
        merge.left=mergeTrees(root1.left,root2.left);
        merge.right=mergeTrees(root1.right,root2.right);
        return merge;
    }

    // 先根遍历
    public void printTree(TreeNode node){
        if(node==null) return;
        System.out.println(node.val);
        if(node.left!=null) printTree(node.left);
        if(node.right!=null)    printTree(node.right);
    }



    public static void main(String[] args) {
//        输入：root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
//        输出：[3,4,5,5,4,null,7]
        Solution solution=new Solution();

        TreeNode node1=new TreeNode(1);
        TreeNode l1=new TreeNode(3);
        node1.left=l1;
        TreeNode r1=new TreeNode(2);
        node1.right=r1;
        l1.left=new TreeNode(5);

        TreeNode node2=new TreeNode(2);
        TreeNode l2=new TreeNode(1);
        node2.left=l2;
        TreeNode r2=new TreeNode(3);
        node2.right=r2;
        l2.right=new TreeNode(4);
        r2.right=new TreeNode(7);
        TreeNode merge=solution.mergeTrees(node1,node2);
        solution.printTree(merge);

    }
}