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
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
{1,3,2,5},{2,1,3,#,4,#,7}
{3,4,5,5,4,#,7}
如题面图
{1},{}
{1}
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) { if (t1 == null) { return t2; } else if (t2 == null) { return t1; } else { t1.val = t1.val + t2.val; t1.left = mergeTrees(t1.left, t2.left); t1.right = mergeTrees(t1.right, t2.right); return t1; } } }
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; } TreeNode root=new TreeNode(t1.val+t2.val); root.left=mergeTrees(t1.left ,t2.left); root.right=mergeTrees(t1.right ,t2.right); return root; }
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 if(t1==null) return t2; if(t2==null) return t1; TreeNode t = new TreeNode (t1.val+t2.val); t.left = mergeTrees(t1.left,t2.left); t.right = mergeTrees(t1.right,t2.right); return t; } }
import java.util.*; //这个算法的空间复杂度难道不是o(n)吗,题目要求的是O(1)啊 /* * 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 //若只有一个节点返回另一个,两个都为null自然返回null if(t1==null){ return t2; } if(t2==null){ return t1; } //根左右的方式递归 TreeNode head=new TreeNode(t1.val+t2.val); head.left=mergeTrees(t1.left,t2.left); head.right=mergeTrees(t1.right,t2.right); return head; } }
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 // 把所加累加的数据放在t1中,如果t1分支节点不存在则将t2中的分支节点挂到t1上 if(t1 == null && t2 == null) { return null; }else if (t1 == null) { return t2; }else 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; } }
public class Solution { /** * * @param t1 TreeNode类 * @param t2 TreeNode类 * @return TreeNode类 */ public TreeNode mergeTrees (TreeNode t1, TreeNode t2) { // 两个节点都为null,则没必要继续比较,直接返回null if(t1 == null && t2 == null) return null; // 拿到两个节点的值之和 int num = (t1 != null ? t1.val : 0) + (t2 != null ? t2.val : 0); if(t1 == null){ // t1等于null时,创建一个对象 存储之和 t1 = new TreeNode(num); }else{ // 否则替换 val t1.val = num; } // 然后继续左子树递归遍历 // 因为将t2 合并到 t1里,所以当递归拿到返回值以后,重新赋值给t1.left t1.left = mergeTrees(t1.left, t2 == null ? null : t2.left); // 然后继续右子树递归遍历 // 因为将t2 合并到 t1里,所以当递归拿到返回值以后,重新赋值给t1.right t1.right = mergeTrees (t1.right, t2 == null ? null : t2.right); // 返回合并结果后的t1 return t1; } }
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 TreeNode t = null; int count = 0; if(t1!=null){ count+=t1.val; } if(t2!=null){ count+=t2.val; } if(count!=0){ t = new TreeNode(count); add(t1.left,t2.left,t,true); add(t1.right,t2.right,t,false); } return t; } public void add(TreeNode t1, TreeNode t2, TreeNode p, boolean flag){ int count = 0; TreeNode t1l = null; TreeNode t1r = null; TreeNode t2l = null; TreeNode t2r = null; if(t1!=null){ count+=t1.val; t1l = t1.left; t1r = t1.right; } if(t2!=null){ count+=t2.val; t2l = t2.left; t2r = t2.right; } if(count!=0){ TreeNode t = new TreeNode(count); if(flag){ p.left = t; }else{ p.right = t; } add(t1l,t2l,t,true); add(t1r,t2r,t,false); } } }
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; TreeNode node=new TreeNode(0); if(t1==null&&t2!=null){ node.val=t2.val; node.left=mergeTrees(null,t2.left); node.right=mergeTrees(null,t2.right); return node; } if(t1!=null&&t2==null){ node.val=t1.val; node.left=mergeTrees(t1.left,null); node.right=mergeTrees(t1.right,null); return node; } node.val=t1.val+t2.val; node.left=mergeTrees(t1.left,t2.left); node.right=mergeTrees(t1.right,t2.right); return node; } }
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 t1 == null ? t2 : t1; TreeNode head = new TreeNode(t1.val + t2.val); head.left = mergeTrees(t1.left, t2.left); head.right = mergeTrees(t1.right, t2.right); return head; } }
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 //都合并到t1上 //本身符合root,子树也符合 if(t1 == null) return t2; if(t2 == null) return t1; //t1和t2都不为空 t1.val = t1.val + t2.val; t1.left = mergeTrees(t1.left, t2.left); t1.right = mergeTrees(t1.right, t2.right); return t1; } }
public TreeNode mergeTrees (TreeNode t1, TreeNode t2) { // write code here 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; }
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 //如果两个节点都为空,则返回false if(t1 ==null && t2 ==null){ return null; }else if(t1 != null && t2 == null){ return t1; }else if(t1 == null && t2 != null){ return t2; }else{ t1.val = t1.val + t2.val; t1.left = mergeTrees(t1.left,t2.left); t1.right = mergeTrees(t1.right,t2.right); return t1; } } }