LeetCode题解： 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.

思路：

和inorder+postorder的思路类似。

题解：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(const vector<int>& preorder, int ps, int pe,
const vector<int>& inorder, int is, int ie)
{
if (is > ie)
return nullptr;

int base = preorder[ps];
TreeNode* node = new TreeNode(base);

int ipos = is;
while(inorder[ipos] != base) ++ipos;

int left_nodes = ipos - is;
int right_nodes = ie - ipos;

if (left_nodes != 0)
node->left = buildTree(preorder, ps + 1, ps + 1 + left_nodes - 1,
inorder, is, is + left_nodes - 1);
if (right_nodes != 0)
node->right = buildTree(preorder, ps + 1 + left_nodes, pe,
inorder, ipos + 1, ie);

return node;
}

TreeNode *buildTree(const vector<int> &preorder,
const vector<int> &inorder) {
if (preorder.empty())
return nullptr;

return buildTree(preorder, 0, preorder.size() - 1,
inorder, 0, inorder.size() - 1);
}
};