- 题目描述
输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
- 题目解析
看到题目想到的是递归求解。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> in) {
int size1 = pre.size();
int size2 = in.size();
if (size1 != size2 || size1 == 0 || size2 == 0)
return NULL;
TreeNode* ret = buildTree(pre, in);
return ret;
}
struct TreeNode* buildTree(vector<int> pre, vector<int> in){
int size = in.size();
TreeNode* ret = new TreeNode(pre[0]);
int lsize = 0, rsize = 0;
vector<int> nextLPre, nextLIn;
vector<int> nextRPre, nextRIn;
bool flag = false;
for (int i = 0; i < size; ++i){
if (in[i] == pre[0]){
flag = true;
}
else if(flag==false){
lsize++;
nextLPre.push_back(pre[i + 1]);
nextLIn.push_back(in[i]);
}
else{
nextRPre.push_back(pre[i]);
nextRIn.push_back(in[i]);
}
}
rsize = size - lsize - 1;
if (lsize != 0)
ret->left = buildTree(nextLPre, nextLIn);
if (rsize != 0)
ret->right = buildTree(nextRPre, nextRIn);
return ret;
}
};