今天是二叉树
层序遍历
AC代码：
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
if (root == nullptr) return 0;
queue<TreeNode*> nodes;
nodes.push(root);
int ans = 0;
TreeNode* node = nullptr;
while (!nodes.empty()) {
int size = nodes.size();
for (int i = 0; i < size; ++i) {
node = nodes.front();
nodes.pop();
if (i == 0) ans = node->val;
if (node->left != nullptr) nodes.push(node->left);
if (node->right != nullptr) nodes.push(node->right);
}
}
return ans;
}
};
//递归
//左节点优先遍历，这样遇到的叶子一定是最左边的
//然后我们再看深度，如果深度更大，那就更新
class Solution {
public:
int maxDepth = INT_MIN;
int result;
void traversal(TreeNode* root, int depth) {
if (root->left == NULL && root->right == NULL) {
if (depth > maxDepth) {
maxDepth = depth;
result = root->val;
}
return;
}
if (root->left) {
traversal(root->left, depth + 1); // 隐藏着回溯
}
if (root->right) {
traversal(root->right, depth + 1); // 隐藏着回溯
}
return;
}
int findBottomLeftValue(TreeNode* root) {
traversal(root, 0);
return result;
}
};
AC代码：
class Solution {
public:
bool hasPathSum(TreeNode* root, int targetSum) {
if(root == nullptr) return false;
if(root->left == nullptr && root->right == nullptr){
if(targetSum - root->val == 0) return true;
return false;
}
return hasPathSum(root->left, targetSum - root->val)
|| hasPathSum(root->right, targetSum - root->val);
}
};
AC代码：
class Solution {
public:
vector<vector<int>> ans;
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
if(root == nullptr) return ans;
GetPath(root, targetSum, vector<int>());
return ans;
}
private:
void GetPath(TreeNode* root, int targetSum, vector<int> path)
{
path.push_back(root->val);
if(root->left == nullptr && root->right == nullptr){
if(targetSum - root->val == 0){
ans.push_back(path);
}
return;
}
if(root->left != nullptr) GetPath(root->left, targetSum - root->val, path);
if(root->right != nullptr) GetPath(root->right, targetSum - root->val, path);
}
};
先从后序遍历中获取根节点的值，也就是postorder的最后一位，然后根据这个值找到它在inorder中的位置，再根据这个位置对inorder进行分割，左边就是左子树，右边就是右子树；然后因为前序遍历的长度必然等于对应的后序遍历的长度，因此接下来我们就可以根据分割后左右inorder子数组的长度对postorder数组进行分割；对于左右子树也是一样的操作，因此不断递归直到传递的数组长度为0
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
int n = inorder.size();
int m = postorder.size();
if(n == 0|| m == 0) return nullptr;
int rootVal = postorder[m-1];
TreeNode* root = new TreeNode(rootVal);
// 获取rootVal在inorder中的位置
int rootValPosInInorder = 0;
while(inorder[rootValPosInInorder] != rootVal){
rootValPosInInorder++;
}
// 按照rootVal将inorder分为两部分
vector<int> leftInorder(inorder.begin(), inorder.begin()+rootValPosInInorder);
vector<int> rightInorder(inorder.begin()+rootValPosInInorder+1, inorder.end());
// 按照inorder分为的两部分的长度将postorder分为两部分
vector<int> leftPostorder(postorder.begin(), postorder.begin()+rootValPosInInorder);
vector<int> rightPostorder(postorder.begin()+rootValPosInInorder, postorder.end()-1);
root->left = buildTree(leftInorder, leftPostorder);
root->right = buildTree(rightInorder, rightPostorder);
return root;
}
};
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return _buildTree(preorder, 0, preorder.size(), inorder, 0, inorder.size());
}
TreeNode* _buildTree(vector<int>& preorder, int preorderBegin, int preorderEnd,
vector<int>& inorder, int inorderBegin, int inorderEnd){
if(preorderBegin == preorderEnd) return nullptr;
int rootVal = preorder[preorderBegin];
TreeNode* root = new TreeNode(rootVal);
//注意我们从0开始，找到的是rootVal在完整的inorder中的下标
//因此之后计算leftInorderEnd时不需要再加上leftInorderBegin
//rootValIndexInInorder本身就是我们要的leftInorderEnd位置
int rootValIndexInInorder = 0;
while(inorder[rootValIndexInInorder] != rootVal){
rootValIndexInInorder++;
}
// 左闭右开
// 先序数组
int leftInorderBegin = inorderBegin;
int leftInorderEnd = rootValIndexInInorder;
int rightInorderBegin = rootValIndexInInorder + 1;
int rightInorderEnd = inorderEnd;
// 后序数组
int leftPreorderBegin = preorderBegin + 1;
int leftPreorderEnd = leftPreorderBegin + (leftInorderEnd - leftInorderBegin);
int rightPreorderBegin = leftPreorderEnd;
int rightPreorderEnd = preorderEnd;
// cout<<leftInorderBegin<<" "<<leftInorderEnd<<" "<<rightInorderBegin<<" "<<rightInorderEnd<<endl;
// cout<<leftPreorderBegin<<" "<<leftPreorderEnd<<" "<<rightPreorderBegin<<" "<<rightPreorderEnd<<endl;
root->left = _buildTree(preorder, leftPreorderBegin, leftPreorderEnd,
inorder, leftInorderBegin, leftInorderEnd);
root->right = _buildTree(preorder, rightPreorderBegin, rightPreorderEnd,
inorder, rightInorderBegin, rightInorderEnd);
return root;
}
};
