105. Construct Binary Tree from Preorder and Inorder Traversal (Tree; DFS)
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
class Solution {
public:
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
root = NULL;
if(inorder.empty()) return root;
root = new TreeNode();
buildSubTree(preorder,inorder,,preorder.size()-,,inorder.size()-,root);
return root; }
void buildSubTree(vector<int> &preorder, vector<int> &inorder, int preStartPos, int preEndPos,int inStartPos, int inEndPos, TreeNode * currentNode) //对于树的递归遍历,如果涉及数组,必定参数中要传递start,end,指明当前子树来源于数组的哪一部分
{
currentNode->val = preorder[preStartPos]; //前序遍历的第一个为根节点 //find root position in inorder vector
int inRootPos;
for(int i = inStartPos; i <= inEndPos; i++)
{
if(inorder[i] == preorder[preStartPos])
{
inRootPos = i;
break;
}
} //分别递归处理左子树和右子树
//left tree:是总序遍历中根节点所在位置之前的部分,对应前序遍历根节点之后相同长度的部分
int newPrePos = preStartPos+max(inRootPos-inStartPos, );
if(inRootPos>inStartPos)
{
currentNode->left = new TreeNode();
buildSubTree(preorder,inorder,preStartPos+, newPrePos,inStartPos,inRootPos-,currentNode->left); //递归调用时,要传递新的start,end
}
//right Tree:是中序遍历根节点所在位置之后的一部分,对应前序遍历从末尾开始相同的长度
if(inRootPos < inEndPos)
{
currentNode->right = new TreeNode();
buildSubTree(preorder,inorder,newPrePos+, preEndPos,inRootPos+,inEndPos,currentNode->right); //递归调用时,要传递新的start,end
} }
private:
TreeNode* root;
};
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