105. Construct Binary Tree from Preorder and Inorder Traversal
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:
/* from Preorder and Inorder Traversal */
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return helper(preorder,0,preorder.size(),inorder,0,inorder.size());
}
TreeNode* helper(vector<int>& preorder,int i,int j,vector<int>& inorder,int ii,int jj)
{
// tree 8 4 5 3 7 3
// preorder 8 [4 3 3 7] [5]
// inorder [3 3 4 7] 8 [5]
// 每次从 preorder 头部取一个值 mid,作为树的根节点
// 检查 mid 在 inorder 中 的位置,则 mid 前面部分将作为 树的左子树,右部分作为树的右子树
if(i >= j || ii >= j)
return NULL;
int mid = preorder[i];
auto f = find(inorder.begin() + ii,inorder.begin() + jj,mid);
int dis = f - inorder.begin() - ii;
TreeNode* root = new TreeNode(mid);
root -> left = helper(preorder,i + 1,i + 1 + dis,inorder,ii,ii + dis);
root -> right = helper(preorder,i + 1 + dis,j,inorder,ii + dis + 1,jj);
return root;
}
};