題目
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
Example 1:
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/ \
4 5
/ \ \
5 4 7
Note: The merging process must start from the root nodes of both trees.
答案:
golang版是自己寫(xiě)的嘁酿，其他的是引用cain_huang 的隙券，文末附上鏈接
golang版，附帶測(cè)試
mergeTrees.go
package _617_mergeTrees
import "fmt"
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
func InitNode(val int, left *TreeNode, right *TreeNode) (ret *TreeNode){
ret = new(TreeNode)
ret.Val = val
ret.Left = left
ret.Right = right
return ret
}
func TreePrint(t1 *TreeNode) {
if nil == t1 {
fmt.Printf("null, ")
return
} else {
fmt.Printf("%+v, ", t1.Val)
}
TreePrint(t1.Left)
TreePrint(t1.Right)
}
func MergeTrees(t1 *TreeNode, t2 *TreeNode) (t3 *TreeNode) {
if nil == t1 && nil == t2 {
return nil
}
if nil != t1 && nil == t2{
return t1
}else if nil == t1 && nil != t2 {
return t2
} else {
t1.Val += t2.Val
t1.Left = MergeTrees(t1.Left, t2.Left)
t1.Right = MergeTrees(t1.Right, t2.Right)
}
return t1
}
mergeTrees_test.go
package _617_mergeTrees
import (
"testing"
"fmt"
)
func treeEqual(t1 *TreeNode, t2 *TreeNode) bool{
if t1.Val != t2.Val {
return false
}
if nil == t1.Left && nil != t2.Left {
return false
}else if nil != t1.Left && nil == t2.Left {
return false
} else if nil != t1.Left && nil != t2.Left {
left := treeEqual(t1.Left, t2.Left)
if !left {
return false
}
}
if nil == t1.Right && nil != t2.Right {
return false
}else if nil != t1.Right && nil == t2.Right {
return false
}else if nil != t1.Right && nil != t2.Right {
right := treeEqual(t1.Right, t2.Right)
if !right {
return false
}
}
return true
}
func Test_mergeTrees(t *testing.T) {
t1l2 := InitNode(5, nil, nil)
t1l1 := InitNode(3, t1l2, nil)
t1r1 := InitNode(2, nil, nil)
t1 := InitNode(1, t1l1, t1r1)
t2l2r1 := InitNode(4, nil, nil)
t2l2r2 := InitNode(7, nil, nil)
t2l1r1 := InitNode(1, nil, t2l2r1)
t2l1r2 := InitNode(3, nil, t2l2r2)
t2 := InitNode(2, t2l1r1, t2l1r2)
t3l2r1 := InitNode(5, nil, nil)
t3l2r2 := InitNode(4, nil, nil)
t3l2r3 := InitNode(7, nil, nil)
t3l1r1 := InitNode(4, t3l2r1, t3l2r2)
t3l1r2 := InitNode(5, nil, t3l2r3)
t3 := InitNode(3, t3l1r1, t3l1r2)
fmt.Printf("merge t1:")
TreePrint(t1)
fmt.Println()
fmt.Printf("merge t2:")
TreePrint(t2)
fmt.Println()
ret := MergeTrees(t1, t2)
ok := treeEqual(t3, ret)
fmt.Printf("merge t3:")
TreePrint(t1)
fmt.Println()
fmt.Printf("t1:%+v, t2:%+v, t3:%+v\n", t1, t2, t3)
if !ok {
t.Errorf("fail, ret.Val want %+v, get %+v\n", t3, ret)
} else {
t.Logf("pass")
}
}
C語(yǔ)言版:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {
if (t1 && t2) {
struct TreeNode *node = (struct TreeNode *)malloc(sizeof(struct TreeNode));
node->val = t1->val + t2->val;
node->left = mergeTrees(t1 ? t1->left : NULL, t2 ? t2->left : NULL);
node->right = mergeTrees(t1 ? t1->right : NULL, t2 ? t2->right : NULL);
return node;
}
return t1 ? t1 : t2;
}
C++版
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
if (t1 && t2) {
TreeNode *node = new TreeNode(t1->val + t2->val);
node->left = mergeTrees(t1 ? t1->left : nullptr, t2 ? t2->left : nullptr);
node->right = mergeTrees(t1 ? t1->right : nullptr, t2 ? t2->right : nullptr);
return node;
}
return t1 ? t1 : t2;
}
};
Java版:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null) return t2;
if (t2 == null) return t1;
int value = (t1 == null ? 0 : t1.val) + (t2 == null ? 0 : t2.val);
TreeNode node = new TreeNode(value);
node.left = mergeTrees(t1 == null ? null : t1.left, t2 == null ? null : t2.left);
node.right = mergeTrees(t1 == null ? null : t1.right, t2 == null ? null : t2.right);
return node;
}
}
鏈接：
