Leetcode 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.

此题目有两种解决思路：

1）递归解决（比较好想）按照手动模拟的思路即可

2）非递归解决，用stack模拟递归

class Solution {
public:
TreeNode *buildTree(vector<int>& preorder, int pre_left,int pre_right,
vector<int>& inorder,  int in_left, int in_right){
if(pre_left > pre_right || in_left > in_right) return NULL;
TreeNode *root = new TreeNode(preorder[pre_left]);
int index = in_left;
for( ; index <= in_right; ++ index ) if(inorder[index] == preorder[pre_left])  break;
int left_cnt = index-in_left;
root->left = buildTree(preorder,pre_left+1,pre_left+left_cnt,inorder,in_left,index-1);
root->right = buildTree(preorder,pre_left+left_cnt+1,pre_right, inorder, index+1,in_right);
return root;
}

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
if(preorder.size() == 0) return NULL;
else return buildTree(preorder,0,preorder.size()-1, inorder,0,inorder.size()-1);
}
};

递归解决