LeetCode OJ:Construct Binary Tree from Preorder and Inorder Traversal

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.

算法思想：

1、先序序列的第一个元素是该树的根；

2、中序序列该元素将该树分为左右两个子树，该元素位置设为i

3、中序序列的0到i-1元素为左子树，对应于先序序列的1到i元素，中序右子树i+1到in_end对应先序的i+1到pre_end元素

所以对于左子树和右子树也满足以上性质，递归即可

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void build(vector<int> &preorder, vector<int> &inorder, TreeNode * &root, int pre_start, int pre_end,int in_start, int in_end){
if(in_start>in_end)return;
root = new TreeNode(preorder[pre_start]);
int i;
for(i = in_start; i<=in_end; i++)
if(inorder[i]==preorder[pre_start])
break;
int pre_mid = pre_start+i-in_start;
build(preorder, inorder, root->left, pre_start+1, pre_mid, in_start, i-1);
build(preorder, inorder, root->right, pre_mid+1, pre_end, i+1, in_end);
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
if(!inorder.size())return NULL;
TreeNode *root = NULL;
build(preorder, inorder, root, 0, preorder.size()-1, 0, inorder.size()-1);
return root;
}
};