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Dec 11, 2023
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feat: add solutions to lc problem: No.2415 #2085

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103 changes: 35 additions & 68 deletions solution/2400-2499/2415.Reverse Odd Levels of Binary Tree/README.md
View file
Original file line number Diff line number Diff line change
@@ -65,9 +65,9 @@

**方法一：BFS**

BFS 遍历二叉树，遍历到奇数层时，反转该层节点的值。
我们可以使用广度优先搜索的方法，用一个队列 $q$ 来存储每一层的节点，用一个变量 $i$ 记录当前层数。若 $i$ 为奇数，则将当前层的节点值反转。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

<!-- tabs:start -->

@@ -87,20 +87,16 @@ class Solution:
q = deque([root])
i = 0
while q:
t = []
if i & 1:
l, r = 0, len(q) - 1
while l < r:
q[l].val, q[r].val = q[r].val, q[l].val
l, r = l + 1, r - 1
for _ in range(len(q)):
node = q.popleft()
if i & 1:
t.append(node)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if t:
j, k = 0, len(t) - 1
while j < k:
t[j].val, t[k].val = t[k].val, t[j].val
j, k = j + 1, k - 1
i += 1
return root
```
@@ -129,30 +125,23 @@ class Solution {
public TreeNode reverseOddLevels(TreeNode root) {
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int i = 0;
while (!q.isEmpty()) {
for (int i = 0; !q.isEmpty(); ++i) {
List<TreeNode> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.pollFirst();
for (int k = q.size(); k > 0; --k) {
var node = q.poll();
if (i % 2 == 1) {
t.add(node);
}
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
if (!t.isEmpty()) {
int j = 0, k = t.size() - 1;
for (; j < k; ++j, --k) {
int v = t.get(j).val;
t.get(j).val = t.get(k).val;
t.get(k).val = v;
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
var x = t.get(l).val;
t.get(l).val = t.get(r).val;
t.get(r).val = x;
}
++i;
}
return root;
}
@@ -177,32 +166,22 @@ class Solution {
public:
TreeNode* reverseOddLevels(TreeNode* root) {
queue<TreeNode*> q{{root}};
int i = 0;
vector<TreeNode*> t;
while (!q.empty()) {
t.clear();
for (int n = q.size(); n; --n) {
for (int i = 0; q.size(); ++i) {
vector<TreeNode*> t;
for (int k = q.size(); k; --k) {
TreeNode* node = q.front();
q.pop();
if (i & 1) {
t.push_back(node);
}
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
if (t.size()) {
int j = 0, k = t.size() - 1;
for (; j < k; ++j, --k) {
int v = t[j]->val;
t[j]->val = t[k]->val;
t[k]->val = v;
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
swap(t[l]->val, t[r]->val);
}
++i;
}
return root;
}
@@ -222,31 +201,22 @@ public:
*/
func reverseOddLevels(root *TreeNode) *TreeNode {
q := []*TreeNode{root}
i := 0
for len(q) > 0 {
for i := 0; len(q) > 0; i++ {
t := []*TreeNode{}
for n := len(q); n > 0; n-- {
for k := len(q); k > 0; k-- {
node := q[0]
q = q[1:]
if i%2 == 1 {
t = append(t, node)
}
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
if len(t) > 0 {
j, k := 0, len(t)-1
for ; j < k; j, k = j+1, k-1 {
v := t[j].Val
t[j].Val = t[k].Val
t[k].Val = v
}
for l, r := 0, len(t)-1; l < r; l, r = l+1, r-1 {
t[l].Val, t[r].Val = t[r].Val, t[l].Val
}
i++
}
return root
}
@@ -270,24 +240,21 @@ func reverseOddLevels(root *TreeNode) *TreeNode {
*/

function reverseOddLevels(root: TreeNode | null): TreeNode | null {
const queue = [root];
let d = 0;
while (queue.length !== 0) {
const n = queue.length;
const t: TreeNode[] = [];
for (let i = 0; i < n; i++) {
const node = queue.shift();
if (d % 2 == 1) {
t.push(node);
const q: TreeNode[] = [root];
for (let i = 0; q.length > 0; ++i) {
if (i % 2) {
for (let l = 0, r = q.length - 1; l < r; ++l, --r) {
[q[l].val, q[r].val] = [q[r].val, q[l].val];
}
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
const m = t.length;
for (let i = 0; i < m >> 1; i++) {
[t[i].val, t[m - 1 - i].val] = [t[m - 1 - i].val, t[i].val];
const nq: TreeNode[] = [];
for (const { left, right } of q) {
if (left) {
nq.push(left);
nq.push(right);
}
}
d++;
q.splice(0, q.length, ...nq);
}
return root;
}
105 changes: 39 additions & 66 deletions solution/2400-2499/2415.Reverse Odd Levels of Binary Tree/README_EN.md
View file
Original file line number Diff line number Diff line change
@@ -58,6 +58,12 @@ The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1

## Solutions

**Solution 1: BFS**

We can use the Breadth-First Search (BFS) method, using a queue $q$ to store the nodes of each level, and a variable $i$ to record the current level. If $i$ is odd, we reverse the values of the nodes at the current level.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

<!-- tabs:start -->

### **Python3**
@@ -74,20 +80,16 @@ class Solution:
q = deque([root])
i = 0
while q:
t = []
if i & 1:
l, r = 0, len(q) - 1
while l < r:
q[l].val, q[r].val = q[r].val, q[l].val
l, r = l + 1, r - 1
for _ in range(len(q)):
node = q.popleft()
if i & 1:
t.append(node)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if t:
j, k = 0, len(t) - 1
while j < k:
t[j].val, t[k].val = t[k].val, t[j].val
j, k = j + 1, k - 1
i += 1
return root
```
@@ -114,30 +116,23 @@ class Solution {
public TreeNode reverseOddLevels(TreeNode root) {
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int i = 0;
while (!q.isEmpty()) {
for (int i = 0; !q.isEmpty(); ++i) {
List<TreeNode> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.pollFirst();
for (int k = q.size(); k > 0; --k) {
var node = q.poll();
if (i % 2 == 1) {
t.add(node);
}
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
if (!t.isEmpty()) {
int j = 0, k = t.size() - 1;
for (; j < k; ++j, --k) {
int v = t.get(j).val;
t.get(j).val = t.get(k).val;
t.get(k).val = v;
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
var x = t.get(l).val;
t.get(l).val = t.get(r).val;
t.get(r).val = x;
}
++i;
}
return root;
}
@@ -162,32 +157,22 @@ class Solution {
public:
TreeNode* reverseOddLevels(TreeNode* root) {
queue<TreeNode*> q{{root}};
int i = 0;
vector<TreeNode*> t;
while (!q.empty()) {
t.clear();
for (int n = q.size(); n; --n) {
for (int i = 0; q.size(); ++i) {
vector<TreeNode*> t;
for (int k = q.size(); k; --k) {
TreeNode* node = q.front();
q.pop();
if (i & 1) {
t.push_back(node);
}
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
if (t.size()) {
int j = 0, k = t.size() - 1;
for (; j < k; ++j, --k) {
int v = t[j]->val;
t[j]->val = t[k]->val;
t[k]->val = v;
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
swap(t[l]->val, t[r]->val);
}
++i;
}
return root;
}
@@ -207,31 +192,22 @@ public:
*/
func reverseOddLevels(root *TreeNode) *TreeNode {
q := []*TreeNode{root}
i := 0
for len(q) > 0 {
for i := 0; len(q) > 0; i++ {
t := []*TreeNode{}
for n := len(q); n > 0; n-- {
for k := len(q); k > 0; k-- {
node := q[0]
q = q[1:]
if i%2 == 1 {
t = append(t, node)
}
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
if len(t) > 0 {
j, k := 0, len(t)-1
for ; j < k; j, k = j+1, k-1 {
v := t[j].Val
t[j].Val = t[k].Val
t[k].Val = v
}
for l, r := 0, len(t)-1; l < r; l, r = l+1, r-1 {
t[l].Val, t[r].Val = t[r].Val, t[l].Val
}
i++
}
return root
}
@@ -255,24 +231,21 @@ func reverseOddLevels(root *TreeNode) *TreeNode {
*/

function reverseOddLevels(root: TreeNode | null): TreeNode | null {
const queue = [root];
let d = 0;
while (queue.length !== 0) {
const n = queue.length;
const t: TreeNode[] = [];
for (let i = 0; i < n; i++) {
const node = queue.shift();
if (d % 2 == 1) {
t.push(node);
const q: TreeNode[] = [root];
for (let i = 0; q.length > 0; ++i) {
if (i % 2) {
for (let l = 0, r = q.length - 1; l < r; ++l, --r) {
[q[l].val, q[r].val] = [q[r].val, q[l].val];
}
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
const m = t.length;
for (let i = 0; i < m >> 1; i++) {
[t[i].val, t[m - 1 - i].val] = [t[m - 1 - i].val, t[i].val];
const nq: TreeNode[] = [];
for (const { left, right } of q) {
if (left) {
nq.push(left);
nq.push(right);
}
}
d++;
q.splice(0, q.length, ...nq);
}
return root;
}
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