Construct Binary Tree from Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
分析: 根据前序遍历和中序遍历构造一棵树，递归求解即可
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void Build(int l1,int l2, int r1,int r2, const vector<int>& pre, const vector<int> & in, TreeNode*& root){
root = new TreeNode(pre[l1]);
int i;
for(i=r1; i<=r2; i++)
if(in[i]==root->val)
break;
if(i==r1)
root->left=nullptr;
else
Build(l1+, l1+i-r1,r1, i- ,pre,in, root->left);
if(i==r2)
root->right=nullptr;
else
Build(l1+i-r1+, l2, i+, r2, pre, in, root->right);
return;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if(preorder.size()== && inorder.size()==)
return nullptr;
TreeNode* root;
Build(,preorder.size()-, , inorder.size()-, preorder, inorder, root);
return root;
}
};