代码随想录day16二叉树4
文章目录
- 513.找树左下角的值
- 112. 路径总和
- 113. 路径总和 II
- 106.从中序与后序遍历序列构造二叉树
- 105. 从前序与中序遍历序列构造二叉树
513.找树左下角的值
题目链接
文章讲解
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
queue<TreeNode*> q;
if(root) q.push(root);
vector<int> res;
int ans=0;
while(!q.empty())
{
int k=q.size();
bool flag=false;
while(k--)
{
TreeNode* node=q.front();
q.pop();
//每次存最左边的值
if(!flag) {ans=node->val;
flag=true;}
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
}
return ans;
}
};
112. 路径总和
题目链接
文章讲解
参照昨天的二叉树的所有路径
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void all(TreeNode* cur,vector<int>& path,int targetSum,bool& flag)
{
path.push_back(cur->val);
if(!cur->left&&!cur->right)
{
int ans=0;
for(int i=0;i<path.size();i++)
{
ans+=path[i];
}
if(ans==targetSum) flag=true;
}
if(cur->left)
{
all(cur->left,path,targetSum,flag);
path.pop_back();
}
if(cur->right)
{
all(cur->right,path,targetSum,flag);
path.pop_back();
}
}
bool hasPathSum(TreeNode* root, int targetSum) {
vector<int> path;
bool flag=false;
if(root==NULL) return false;
all(root,path,targetSum,flag);
return flag;
}
};
113. 路径总和 II
题目链接
文章讲解
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void all(TreeNode* cur,vector<int>& path,int targetSum,vector<vector<int>>& res)
{
path.push_back(cur->val);
if(!cur->left&&!cur->right)
{
int ans=0;
for(int i=0;i<path.size();i++)
{
ans+=path[i];
}
if(ans==targetSum)
{
res.push_back(path);
}
}
if(cur->left)
{
all(cur->left,path,targetSum,res);
path.pop_back();
}
if(cur->right)
{
all(cur->right,path,targetSum,res);
path.pop_back();
}
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
vector<int> path;
vector<vector<int>> res;
if(root==NULL) return res;
all(root,path,targetSum,res);
return res;
}
};
106.从中序与后序遍历序列构造二叉树
题目链接
文章讲解
第一步：如果数组大小为零的话，说明是空节点了。
第二步：如果不为空，那么取后序数组最后一个元素作为节点元素。
第三步：找到后序数组最后一个元素在中序数组的位置，作为切割点
第四步：切割中序数组，切成中序左数组和中序右数组 （顺序别搞反了，一定是先切中序数组）
第五步：切割后序数组，切成后序左数组和后序右数组
第六步：递归处理左区间和右区间
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* solve(vector<int>& inorder, vector<int>& postorder)
{
if(postorder.size()==0) return NULL;
int fenge=postorder[postorder.size()-1];
TreeNode* node=new TreeNode(fenge);
if(postorder.size()==1) return node;
int k=0;
for(int i=0;i<inorder.size();i++)
{
if(inorder[i]==fenge)
{
k=i;
break;
}
}
vector<int> leftinorder(inorder.begin(),inorder.begin()+k);
//特别注意这里要加1
vector<int> rightinorder(inorder.begin()+k+1,inorder.end());
postorder.pop_back();
vector<int> leftpostorder(postorder.begin(),postorder.begin()+leftinorder.size());
vector<int> rightpostorder(postorder.begin()+leftinorder.size(),postorder.end());
node->left=solve(leftinorder,leftpostorder);
node->right=solve(rightinorder,rightpostorder);
return node;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
if(inorder.size()==0||postorder.size()==0) return NULL;
return solve(inorder,postorder);
}
};
105. 从前序与中序遍历序列构造二叉树
题目链接
文章讲解
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* solve(vector<int>& preorder, vector<int>& inorder)
{
if(preorder.size()==0) return NULL;
int fenge=preorder[0];
TreeNode* node=new TreeNode(fenge);
if(preorder.size()==1) return node;
int k;
for(int i=0;i<inorder.size();i++)
{
if(inorder[i]==fenge)
{
k=i;
break;
}
}
vector<int> leftinorder(inorder.begin(),inorder.begin()+k);
vector<int> rightinorder(inorder.begin()+k+1,inorder.end());
preorder.erase(preorder.begin());
vector<int> leftpreorder(preorder.begin(),preorder.begin()+leftinorder.size());
vector<int> rightpreorder(preorder.begin()+leftinorder.size(),preorder.end());
node->left=solve(leftpreorder,leftinorder);
node->right=solve(rightpreorder,rightinorder);
return node;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if(preorder.size()==0||inorder.size()==0) return NULL;
return solve(preorder,inorder);
}
};