95. Unique Binary Search Trees II (Tree; DFS)
Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
For example,
Given n = 3, your program should return all 5 unique BST's shown below.
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
思路:结果要保留树的所有节点,所以每次都new出新节点。
new新节点的顺序是从下往上(否则在遍历到子节点的时候,还得深度拷贝所有的父辈节点),所以要先访问左、右儿子,再访问根节点=>后序遍历。
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
vector<TreeNode*> result;
if(n==)
{
return result;
}
postOrderTraverse(,n, result);
return result;
} void postOrderTraverse(int start, int end, vector<TreeNode*> &rootArray)
{
if(start == end){ //递归结束条件:碰到叶子节点(没有子节点)
rootArray.push_back(new TreeNode(start));
return;
} for(int i = start; i<=end; i++){ //iterate all roots
vector<TreeNode*> leftTree;
vector<TreeNode*> rightTree;
if(i > start){ //build left tree
postOrderTraverse(start, i-, leftTree);
}
if(i < end){ //build right tree
postOrderTraverse(i+, end, rightTree);
} //visit root: build a new tree for each (leftTree, rightTree) pair
if(leftTree.empty()){
for(int j = ; j< rightTree.size(); j++){
TreeNode* root = new TreeNode(i);
root->right = rightTree[j];
rootArray.push_back(root);
}
}
else if(rightTree.empty()){
for(int j = ; j< leftTree.size(); j++){
TreeNode* root = new TreeNode(i);
root->left = leftTree[j];
rootArray.push_back(root);
}
}
else{
for(int j = ; j< leftTree.size(); j++)
{
for(int k = ; k< rightTree.size(); k++)
{
TreeNode* root = new TreeNode(i);
root->left = leftTree[j];
root->right = rightTree[k];
rootArray.push_back(root);
}
}
}
}
}
};
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