N叉树的前序遍历
地址:https://leetcode-cn.com/problems/n-ary-tree-preorder-traversal/
给定一个 n 叉树的根节点 root ,返回 其节点值的 前序遍历 。
n 叉树 在输入中按层序遍历进行序列化表示,每组子节点由空值 null 分隔(请参见示例)。
示例 1:
输入:root = [1,null,3,2,4,null,5,6]
输出:[1,3,5,6,2,4]
示例 2:
输入:root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出:[1,2,3,6,7,11,14,4,8,12,5,9,13,10]
提示:
节点总数在范围 [0, 104]内
0 <= Node.val <= 104
n 叉树的高度小于或等于 1000
进阶:递归法很简单,你可以使用迭代法完成此题吗?
使用OC 我们需要创建一个BinaryTreeNode节点model 里面的对象如下
/**
* 值
*/
@property (nonatomic, assign) NSInteger value;
/**
* N节点
*/
@property (nonatomic, strong) NSArray <BinaryTreeNode *> *dataArray;
递归
很简单 上代码吧
- (NSArray *)preorder1:(BinaryTreeNode *)root
{
if (root == nil) {
return @[];
}
NSMutableArray *values = [NSMutableArray array];
[values addObject:@(root.value)];
for (BinaryTreeNode *node in root.dataArray) {
NSArray *dataArray = [self preorder1:node];
[values addObjectsFromArray:dataArray];
}
return values;
}
迭代
- (NSArray *)preorder2:(BinaryTreeNode *)root
{
if (root == nil) {
return @[];
}
NSMutableArray *values = [NSMutableArray array];
NSMutableArray *dataArray = [NSMutableArray array];
[dataArray addObject:root];
while (dataArray.count > 0) {
BinaryTreeNode *node = [dataArray objectAtIndex:0];
[dataArray removeObjectAtIndex:0];
[values addObject:@(node.value)];
for (NSInteger i=node.dataArray.count; i>0; i--) {
BinaryTreeNode *node1 = node.dataArray[i-1];
[dataArray insertObject:node1 atIndex:0];
}
}
return values;
}
验证
{
/*
输入:root = [1,null,3,2,4,null,5,6]
输出:[1,3,5,6,2,4]
*/
BinaryTreeNode *node1 = [[BinaryTreeNode alloc] init];
node1.value = 1;
BinaryTreeNode *node2 = [[BinaryTreeNode alloc] init];
node2.value = 3;
BinaryTreeNode *node3 = [[BinaryTreeNode alloc] init];
node3.value = 2;
BinaryTreeNode *node4 = [[BinaryTreeNode alloc] init];
node4.value = 4;
node1.dataArray = @[node2, node3, node4];
BinaryTreeNode *node5 = [[BinaryTreeNode alloc] init];
node5.value = 5;
BinaryTreeNode *node6 = [[BinaryTreeNode alloc] init];
node6.value = 6;
node2.dataArray = @[node5, node6];
NSArray *arr1 = [self preorder1:node1];
NSLog(@"arr1 is %@",arr1);
NSArray *arr2 = [self preorder2:node1];
NSLog(@"arr2 is %@",arr2);
}
{
/*
输入:root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出:[1,2,3,6,7,11,14,4,8,12,5,9,13,10]
*/
BinaryTreeNode *node1 = [[BinaryTreeNode alloc] init];
node1.value = 1;
BinaryTreeNode *node2 = [[BinaryTreeNode alloc] init];
node2.value = 2;
BinaryTreeNode *node3 = [[BinaryTreeNode alloc] init];
node3.value = 3;
BinaryTreeNode *node4 = [[BinaryTreeNode alloc] init];
node4.value = 4;
BinaryTreeNode *node5 = [[BinaryTreeNode alloc] init];
node5.value = 5;
node1.dataArray = @[node2, node3, node4, node5];
BinaryTreeNode *node6 = [[BinaryTreeNode alloc] init];
node6.value = 6;
BinaryTreeNode *node7 = [[BinaryTreeNode alloc] init];
node7.value = 7;
node3.dataArray = @[node6, node7];
BinaryTreeNode *node8 = [[BinaryTreeNode alloc] init];
node8.value = 8;
node4.dataArray = @[node8];
BinaryTreeNode *node9 = [[BinaryTreeNode alloc] init];
node9.value = 9;
BinaryTreeNode *node10 = [[BinaryTreeNode alloc] init];
node10.value = 10;
node5.dataArray = @[node9, node10];
BinaryTreeNode *node11 = [[BinaryTreeNode alloc] init];
node11.value = 11;
node7.dataArray = @[node11];
BinaryTreeNode *node12 = [[BinaryTreeNode alloc] init];
node12.value = 12;
node8.dataArray = @[node12];
BinaryTreeNode *node13 = [[BinaryTreeNode alloc] init];
node13.value = 13;
node9.dataArray = @[node13];
BinaryTreeNode *node14 = [[BinaryTreeNode alloc] init];
node14.value = 14;
node11.dataArray = @[node14];
NSArray *arr1 = [self preorder1:node1];
NSLog(@"arr1 is %@",arr1);
NSArray *arr2 = [self preorder2:node1];
NSLog(@"arr2 is %@",arr2);
}
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