已知两颗二叉树，将它们合并成一颗二叉树。合并规则是：都存在的结点，就将结点值加起来，否则空的位置就由另一个树的结点来代替。例如： 两颗二叉树是: Tree 1 Tree 2 合并后的树为 数据范围：树上节点数量满足 ，树上节点的值一定在32位整型范围内。 进阶：空间复杂度 ，时间复杂度
示例1
输入
{1,3,2,5},{2,1,3,#,4,#,7}
输出
{3,4,5,5,4,#,7}
说明
如题面图
示例2
输入
{1},{}
输出
{1}
加载中...
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ public TreeNode mergeTrees (TreeNode t1, TreeNode t2) { // write code here } }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { // write code here } };
#coding:utf-8 # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param t1 TreeNode类 # @param t2 TreeNode类 # @return TreeNode类 # class Solution: def mergeTrees(self , t1 , t2 ): # write code here
using System; using System.Collections.Generic; /* public class TreeNode { public int val; public TreeNode left; public TreeNode right; public TreeNode (int x) { val = x; } } */ class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ public TreeNode mergeTrees (TreeNode t1, TreeNode t2) { // write code here } }
/* * function TreeNode(x) { * this.val = x; * this.left = null; * this.right = null; * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ function mergeTrees( t1 , t2 ) { // write code here } module.exports = { mergeTrees : mergeTrees };
<?php /*class TreeNode{ var $val; var $left = NULL; var $right = NULL; function __construct($val){ $this->val = $val; } }*/ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ function mergeTrees( $t1 , $t2 ) { // write code here }
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param t1 TreeNode类 # @param t2 TreeNode类 # @return TreeNode类 # class Solution: def mergeTrees(self , t1: TreeNode, t2: TreeNode) -> TreeNode: # write code here
package main import "fmt" import . "nc_tools" /* * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ func mergeTrees( t1 *TreeNode , t2 *TreeNode ) *TreeNode { // write code here }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2 ) { // write code here }
# class TreeNode # attr_accessor :val, :left, :right # def initialize(val, left = nil, right = nil) # @val, @left, @right = val, left, right # end # end # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param t1 TreeNode类 # @param t2 TreeNode类 # @return TreeNode类 # class Solution def mergeTrees(t1, t2) # write code here end end
/** * class TreeNode(var val: Int) { * var left: TreeNode = null * var right: TreeNode = null * } */ object Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ def mergeTrees(t1: TreeNode,t2: TreeNode): TreeNode = { // write code here } }
/** * class TreeNode(var `val`: Int) { * var left: TreeNode? = null * var right: TreeNode? = null * } */ object Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ fun mergeTrees(t1: TreeNode?,t2: TreeNode?): TreeNode? { // write code here } }
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ public TreeNode mergeTrees (TreeNode t1, TreeNode t2) { // write code here } }
/*class TreeNode { * val: number * left: TreeNode | null * right: TreeNode | null * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ export function mergeTrees(t1: TreeNode, t2: TreeNode): TreeNode { // write code here }
/** * public class TreeNode { * public var val: Int * public var left: TreeNode? * public var right: TreeNode? * public init(_ val: Int=0, _ left: TreeNode?=nil, _ right: TreeNode?=nil) { * self.val = val * self.left = left * self.right = right * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ func mergeTrees ( _ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? { // write code here } }
/** * #[derive(PartialEq, Eq, Debug, Clone)] * pub struct TreeNode { * pub val: i32, * pub left: Option<Box<TreeNode>>, * pub right: Option<Box<TreeNode>>, * } * * impl TreeNode { * #[inline] * fn new(val: i32) -> Self { * TreeNode { * val: val, * left: None, * right: None, * } * } * } */ struct Solution{ } impl Solution { fn new() -> Self { Solution{} } /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ pub fn mergeTrees(&self, t1: Option<Box<TreeNode>>, t2: Option<Box<TreeNode>>) -> Option<Box<TreeNode>> { // write code here } }
{1,3,2,5},{2,1,3,#,4,#,7}
{3,4,5,5,4,#,7}
{1},{}
{1}