代码随想录算法训练营第十八天|● 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:513. 找树左下角的值 - 力扣(LeetCode)
思路:利用层序遍历,找到最后一层在输出第一个节点
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
queue<TreeNode*>q;
int result;
q.push(root);
q.push(NULL);
TreeNode*node=root;
result=root->val;
while(!q.empty()){
if(q.front()==NULL){
q.pop();
if(!q.empty())
result=q.front()->val;
else break;
q.push(NULL);
}
node=q.front();
q.pop();
if(node->left){q.push(node->left);}
if(node->right){q.push(node->right);}
}
return result;
}
};
112路径总和
class Solution {
public:
bool traversal(TreeNode *cur,int count){
if(!cur->right&&!cur->left&&count)return false;
if(!cur->right&&!cur->left&&!count)return true;
if(cur->left){
count-=cur->left->val;
if(traversal(cur->left,count))return true;
count+=cur->left->val;
}
if(cur->right){
count-=cur->right->val;
if(traversal(cur->right,count))return true;
count+=cur->right->val;
}
return false;
}
bool hasPathSum(TreeNode* root, int targetSum) {
if(root==NULL)return false;
return traversal(root,targetSum-root->val);
}
};
106.从中序与后序遍历序列构造二叉树
题目链接:106. 从中序与后序遍历序列构造二叉树 - 力扣(LeetCode)
本题投降
class Solution {
private:
// 中序区间:[inorderBegin, inorderEnd),后序区间[postorderBegin, postorderEnd)
TreeNode* traversal (vector<int>& inorder, int inorderBegin, int inorderEnd, vector<int>& postorder, int postorderBegin, int postorderEnd) {
if (postorderBegin == postorderEnd) return NULL;
int rootValue = postorder[postorderEnd - 1];
TreeNode* root = new TreeNode(rootValue);
if (postorderEnd - postorderBegin == 1) return root;
int delimiterIndex;
for (delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; delimiterIndex++) {
if (inorder[delimiterIndex] == rootValue) break;
}
// 切割中序数组
// 左中序区间,左闭右开[leftInorderBegin, leftInorderEnd)
int leftInorderBegin = inorderBegin;
int leftInorderEnd = delimiterIndex;
// 右中序区间,左闭右开[rightInorderBegin, rightInorderEnd)
int rightInorderBegin = delimiterIndex + 1;
int rightInorderEnd = inorderEnd;
// 切割后序数组
// 左后序区间,左闭右开[leftPostorderBegin, leftPostorderEnd)
int leftPostorderBegin = postorderBegin;
int leftPostorderEnd = postorderBegin + delimiterIndex - inorderBegin; // 终止位置是 需要加上 中序区间的大小size
// 右后序区间,左闭右开[rightPostorderBegin, rightPostorderEnd)
int rightPostorderBegin = postorderBegin + (delimiterIndex - inorderBegin);
int rightPostorderEnd = postorderEnd - 1; // 排除最后一个元素,已经作为节点了
root->left = traversal(inorder, leftInorderBegin, leftInorderEnd, postorder, leftPostorderBegin, leftPostorderEnd);
root->right = traversal(inorder, rightInorderBegin, rightInorderEnd, postorder, rightPostorderBegin, rightPostorderEnd);
return root;
}
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
if (inorder.size() == 0 || postorder.size() == 0) return NULL;
// 左闭右开的原则
return traversal(inorder, 0, inorder.size(), postorder, 0, postorder.size());
}
};