文章目录
- 1、 [根据二叉树创建字符串](https://leetcode.cn/problems/construct-string-from-binary-tree/)
- 2、 [二叉树的层序遍历](https://leetcode.cn/problems/binary-tree-level-order-traversal/)
- 3、 [二叉树的层序遍历 II](https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/)
- 4、 [二叉树的最近公共祖先](https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/)
- 5、 [二叉搜索树与双向链表](https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/)
- 6、 [从前序与中序遍历序列构造二叉树](https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
- 7、 [从中序与后序遍历序列构造二叉树](https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)
- 8、 [二叉树的前序遍历-非递归](https://leetcode.cn/problems/binary-tree-preorder-traversal/)
- 9、 [二叉树的中序遍历-非递归](https://leetcode.cn/problems/binary-tree-inorder-traversal/)
- 10、 [二叉树的后序遍历-非递归](https://leetcode.cn/problems/binary-tree-postorder-traversal/)
1、 根据二叉树创建字符串
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思路图解:
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string tree2str(TreeNode* root)
{
if(root == nullptr)
return "";
string str = to_string(root->val);
if(root->left || root->right)
{
str += '(';
str += tree2str(root->left);
str += ')';
}
if(root->right)
{
str += '(';
str += tree2str(root->right);
str += ')';
}
return str;
}
2、 二叉树的层序遍历
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vector<vector<int>> levelOrder(TreeNode* root)
{
vector<vector<int>> vv;
if(root == nullptr)
return vv;
queue<TreeNode*> Qdata;
Qdata.push(root);
int levelSize = 1;
while(!Qdata.empty())
{
vector<int> v;
while(levelSize--)
{
TreeNode* front = Qdata.front();
Qdata.pop();
v.push_back(front->val);
if(front->left)
Qdata.push(front->left);
if(front->right)
Qdata.push(front->right);
}
vv.push_back(v);
levelSize = Qdata.size();
}
return vv;
}
3、 二叉树的层序遍历 II
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思路:直接将从头层序遍历的结果倒过来即可
vector<vector<int>> levelOrderBottom(TreeNode* root)
{
vector<vector<int>> vv;
queue<TreeNode*> q1;
int levelSize = 0;
if(root != nullptr)
{
q1.push(root);
levelSize = 1;
}
while(!q1.empty())
{
vector<int> v;
while(levelSize--)
{
TreeNode* front = q1.front();
q1.pop();
v.push_back(front->val);
if(front->left)
q1.push(front->left);
if(front->right)
q1.push(front->right);
}
levelSize = q1.size();
vv.push_back(v);
}
reverse(vv.begin(),vv.end());
return vv;
}
4、 二叉树的最近公共祖先
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bool GetPath(TreeNode* root, TreeNode* x,stack<TreeNode*>& path)
{
if(root == nullptr)
return false;
path.push(root);
if(root == x)
return true;
bool Inleft = GetPath(root->left, x,path);
if(Inleft)
return true;
bool Inright = GetPath(root->right, x,path);
if(Inright)
return true;
path.pop();
return false;
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
stack<TreeNode*> ppath;
stack<TreeNode*> qpath;
GetPath(root,p,ppath);
GetPath(root,q,qpath);
while(ppath.size() != qpath.size())
{
if(ppath.size() > qpath.size())
ppath.pop();
else
{
qpath.pop();
}
}
while(ppath.top() != qpath.top())
{
ppath.pop();
qpath.pop();
}
return ppath.top();
}
5、 二叉搜索树与双向链表
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void InOrder(TreeNode* cur,TreeNode*& prev)
{
if(cur == nullptr)
return;
InOrder(cur->left,prev);
cur->left = prev;
if(prev)
prev->right = cur;
prev = cur;
InOrder(cur->right,prev);
}
TreeNode* Convert(TreeNode* pRootOfTree)
{
if(pRootOfTree == nullptr)
return nullptr;
TreeNode* prev = nullptr;
InOrder(pRootOfTree,prev);
TreeNode* head = pRootOfTree;
while(head && head->left)
{
head = head->left;
}
return head;
}
6、 从前序与中序遍历序列构造二叉树
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TreeNode* _buildTree(vector<int>& preorder, vector<int>& inorder,int& prei,int inbegin,int inend)
{
if(inbegin > inend)
return nullptr;
TreeNode* root = new TreeNode(preorder[prei]);
int rooti = inbegin;
while(inbegin <= inend)
{
if(preorder[prei] == inorder[rooti])
break;
else
++rooti;
}
++prei;
root->left = _buildTree(preorder, inorder,prei,inbegin,rooti-1);
root->right = _buildTree(preorder, inorder,prei,rooti+1,inend);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
{
int i = 0;
return _buildTree(preorder,inorder,i,0,inorder.size()-1);
}
7、 从中序与后序遍历序列构造二叉树
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TreeNode* _buildTree(vector<int>& inorder, vector<int>& postorder,int& posti,int inbegin,int inend)
{
if(inbegin > inend)
return nullptr;
TreeNode* root = new TreeNode(postorder[posti]);
int rooti = inbegin;
while(rooti <= inend)
{
if(inorder[rooti] == postorder[posti])
break;
else
rooti++;
}
posti--;
root->right = _buildTree(inorder, postorder,posti,rooti+1,inend);
root->left = _buildTree(inorder, postorder,posti,inbegin,rooti - 1);
return root;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
{
int i = postorder.size() - 1;
return _buildTree(inorder, postorder,i,0,inorder.size() - 1);
}
8、 二叉树的前序遍历-非递归
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vector<int> preorderTraversal(TreeNode* root)
{
stack<TreeNode*> st;
vector<int> v;
TreeNode* cur = root;
while(cur || !st.empty())
{
while(cur)
{
v.push_back(cur->val);
st.push(cur);
cur = cur->left;
}
TreeNode* top = st.top();
st.pop();
cur = top->right;
}
return v;
}
9、 二叉树的中序遍历-非递归
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vector<int> inorderTraversal(TreeNode* root)
{
stack<TreeNode*> st;
vector<int> v;
TreeNode* cur = root;
while(cur || !st.empty())
{
while(cur)
{
st.push(cur);
cur = cur->left;
}
TreeNode* top = st.top();
st.pop();
v.push_back(top->val);
cur = top->right;
}
return v;
}
10、 二叉树的后序遍历-非递归
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vector<int> postorderTraversal(TreeNode* root)
{
stack<TreeNode*> st;
vector<int> v;
TreeNode* cur = root;
TreeNode* prev = nullptr;
while(cur || !st.empty())
{
while(cur)
{
st.push(cur);
cur = cur->left;
}
TreeNode* top = st.top();
if(top->right == nullptr || top->right == prev)
{
v.push_back(top->val);
st.pop();
prev = top;
}
else
{
cur = top->right;
}
}
return v;
}