class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
queueque;
que.push(root);
int res=0;
while(!que.empty()){
int n=que.size();
int mid=que.front()->val;
for(int i=0;ileft)
que.push(node->left);
if(node->right)
que.push(node->right);
}
if(que.empty())
res=mid;
}
return res;
}
};
class Solution {
public:
vector> ans;
void judge(TreeNode*root,int deepth)
{
if(root==nullptr)
return;
if(root->left==nullptr && root->right==nullptr)
ans.push_back({deepth,root->val});
judge(root->left,deepth+1);
judge(root->right,deepth+1);
}
int findBottomLeftValue(TreeNode* root) {
judge(root,0);
int mid=0,res=0;
for(int i=0;imid){
mid=ans[i][0];
res=i;
}
}
return ans[res][1];
}
};
class Solution {
public:
int maxDeepth=-1;
int res=0;
void judge(TreeNode*root,int deepth)
{
if(root==nullptr)
return;
if(root->left==nullptr && root->right==nullptr)//遍历到叶子节点时
{
if(deepth>maxDeepth){
maxDeepth=deepth;
res=root->val;
}
}
judge(root->left,deepth+1);
judge(root->right,deepth+1);
}
int findBottomLeftValue(TreeNode* root) {
judge(root,0);
return res;
}
};
class Solution {
public:
bool res=false;
void judge(TreeNode*root,int targetSum,int sum)
{
if(root==nullptr)
return;
if(root->left==nullptr && root->right==nullptr){
if(sum+root->val==targetSum)
res=true;
//cout<val;
// if(sum>targetSum)
// return;
judge(root->left,targetSum,sum);
judge(root->right,targetSum,sum);
}
bool hasPathSum(TreeNode* root, int targetSum) {
judge(root,targetSum,0);
return res;
}
};
class Solution {
public:
vector>res;
vectormids;
void judge(TreeNode*root,int targetSum,int sum){
if(root==nullptr)
return;
if(root->left==nullptr && root->right==nullptr){
if(sum+root->val==targetSum){
mids.push_back(root->val);
res.push_back(mids);
mids.erase(mids.end()-1);
return;
}
}
sum+=root->val;
mids.push_back(root->val);
judge(root->left,targetSum,sum);
judge(root->right,targetSum,sum);
mids.erase(mids.end()-1);
}
vector> pathSum(TreeNode* root, int targetSum) {
//思路一:直接递归,在叶子节点处判断
if(root==nullptr)
return vector>();
judge(root,targetSum,0);
return res;
}
};
class Solution {
private:
TreeNode*judge(vector&inorder,vector&postorder){
if(postorder.size()==0) return NULL;
//后续遍历数组最后一个元素,就是当前二叉树的中间节点
int rootValue=postorder[postorder.size()-1];
TreeNode*root=new TreeNode(rootValue);
//叶子节点
if(postorder.size()==1) return root;
//找到中序遍历的切割点
int midLastIndex;
for(midLastIndex=0;midLastIndex leftInorder(inorder.begin(),inorder.begin()+midLastIndex);
vector rightInorder(inorder.begin()+midLastIndex+1,inorder.end());
//删除后序数组的末尾元素
postorder.resize(postorder.size()-1);
//切割后序数组
//依然左闭右开,注意这里使用了左中序数组大小作为切割点
vector leftPostorder(postorder.begin(),postorder.begin()+leftInorder.size());
vector rightPostorder(postorder.begin()+leftInorder.size(),postorder.end());
root->left=judge(leftInorder,leftPostorder);
root->right=judge(rightInorder,rightPostorder);
return root;
}
public:
TreeNode* buildTree(vector& inorder, vector& postorder) {
if(inorder.empty() || postorder.empty()) return nullptr;
return judge(inorder,postorder);
}
};
今日问题:43.字符串相乘
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