目录
1. 从中序与后序遍历序列构造二叉树 🌟🌟
2. 从先序与中序遍历序列构造二叉树 🌟🌟
3. 二叉树展开为链表 🌟🌟
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1. 从中序与后序遍历序列构造二叉树
根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
出处：
https://edu.csdn.net/practice/26654112
代码：
#define null INT_MIN
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
{
if (0 == inorder.size() || 0 == postorder.size())
{
return NULL;
}
return build(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
}
TreeNode *build(vector<int> &inorder, int i1, int i2, vector<int> &postorder, int p1, int p2)
{
TreeNode *root = new TreeNode(postorder[p2]);
int i = i1;
while (i <= i2 && postorder[p2] != inorder[i])
{
i++;
}
int left = i - i1;
int right = i2 - i;
if (left > 0)
{
root->left = build(inorder, i1, i - 1, postorder, p1, p1 + left - 1);
}
if (right > 0)
{
root->right = build(inorder, i + 1, i2, postorder, p1 + left, p2 - 1);
}
return root;
}
};
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
st.pop();
if (node != nullptr) {
res.push_back(node->val);
st.push(node->right);
st.push(node->left);
}
}
return res;
}
string vectorToString(vector<int> vect) {
stringstream ss;
ss << "[";
for (size_t i = 0; i < vect.size(); i++) {
ss << vect[i] << (i < vect.size() - 1 ? "," : "");
}
ss << "]";
return ss.str();
}
int main()
{
Solution s;
vector<int> inorder = {9,3,15,20,7};
vector<int> postorder = {9,15,7,20,3};
TreeNode* root = s.buildTree(inorder, postorder);
vector<int> preorder = preorderTraversal(root);
cout << vectorToString(preorder) << endl;
return 0;
}
输出：
[3,9,20,15,7]
2. 从先序与中序遍历序列构造二叉树
根据一棵树的先序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
先序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
出处：
与上题类似，已知三种遍历中的两个且有中序，就能确定一棵二叉树。如果只有先序、后序，因为不能确定根的位置，所以无法确定。
代码：
#define null INT_MIN
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty() || inorder.empty()) {
return nullptr;
}
unordered_map<int, int> pos;
for (int i = 0; i < (int)inorder.size(); ++i) {
pos[inorder[i]] = i;
}
return build(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1, pos);
}
TreeNode* build(vector<int>& preorder, int p1, int p2, vector<int>& inorder, int i1, int i2, unordered_map<int, int>& pos) {
if (p1 > p2 || i1 > i2) {
return nullptr;
}
int rootVal = preorder[p1];
int i = pos[rootVal];
TreeNode* root = new TreeNode(rootVal);
root->left = build(preorder, p1 + 1, p1 + i - i1, inorder, i1, i - 1, pos);
root->right = build(preorder, p1 + i - i1 + 1, p2, inorder, i + 1, i2, pos);
return root;
}
};
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
if (root == nullptr) {
return res;
}
stack<TreeNode*> st1, st2;
st1.push(root);
while (!st1.empty()) {
TreeNode* node = st1.top();
st1.pop();
st2.push(node);
if (node->left != nullptr) {
st1.push(node->left);
}
if (node->right != nullptr) {
st1.push(node->right);
}
}
while (!st2.empty()) {
res.push_back(st2.top()->val);
st2.pop();
}
return res;
}
string vectorToString(vector<int> vect) {
stringstream ss;
ss << "[";
for (size_t i = 0; i < vect.size(); i++) {
ss << vect[i] << (i < vect.size() - 1 ? "," : "");
}
ss << "]";
return ss.str();
}
int main()
{
Solution s;
vector<int> preorder = {3,9,20,15,7};
vector<int> inorder = {9,3,15,20,7};
TreeNode* root = s.buildTree(preorder, inorder);
vector<int> postorder = postorderTraversal(root);
cout << vectorToString(postorder) << endl;
return 0;
}
输出：
[9,15,7,20,3]
3. 二叉树展开为链表
给你二叉树的根结点 root ，请你将它展开为一个单链表：
- 展开后的单链表应该同样使用
TreeNode，其中right子指针指向链表中下一个结点，而左子指针始终为null。 - 展开后的单链表应该与二叉树 先序遍历 顺序相同。
示例 1：
输入：root = [1,2,5,3,4,null,6] 输出：[1,null,2,null,3,null,4,null,5,null,6]
示例 2：
输入：root = [] 输出：[]
示例 3：
输入：root = [0] 输出：[0]
提示：
- 树中结点数在范围
[0, 2000]内 -100 <= Node.val <= 100
进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？
出处：
https://edu.csdn.net/practice/25405424
代码：
#define null INT_MIN
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
void rconnect(TreeNode *&node, TreeNode *pmove)
{
if (pmove == nullptr)
return;
node->right = new TreeNode(pmove->val);
node->left = nullptr;
node = node->right;
rconnect(node, pmove->left);
rconnect(node, pmove->right);
}
void flatten(TreeNode *root)
{
if (root == nullptr)
return;
TreeNode *head = new TreeNode(null);
TreeNode *newroot = head;
rconnect(head, root);
newroot = newroot->right->right;
root->right = newroot;
root->left = nullptr;
}
};
TreeNode* buildTree(vector<int>& nums)
{
if (nums.empty()) return nullptr;
TreeNode *root = new TreeNode(nums.front());
queue<TreeNode*> q;
q.push(root);
int i = 1;
while(!q.empty() && i < nums.size())
{
TreeNode *cur = q.front();
q.pop();
if(i < nums.size() && nums[i] != null)
{
cur->left = new TreeNode(nums[i]);
q.push(cur->left);
}
i++;
if(i < nums.size() && nums[i] != null)
{
cur->right = new TreeNode(nums[i]);
q.push(cur->right);
}
i++;
}
return root;
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
st.pop();
if (node != nullptr) {
res.push_back(node->val);
st.push(node->right);
st.push(node->left);
}
else
res.push_back(null);
}
while (res.back()==null)
res.pop_back();
return res;
}
string vectorToString(vector<int> vect) {
stringstream ss;
ss << "[";
for (size_t i = 0; i < vect.size(); i++)
{
ss << (vect[i] == null ? "null" : to_string(vect[i]));
ss << (i < vect.size() - 1 ? "," : "]");
}
return ss.str();
}
int main() {
Solution s;
vector<int> nums = {1,2,5,3,4,null,6};
TreeNode* root = buildTree(nums);
s.flatten(root);
cout << vectorToString(preorderTraversal(root)) << endl;
return 0;
}输出：
[1,null,2,null,3,null,4,null,5,null,6]
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