[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal

[Problem]
Given preorder and inorder traversal of a tree, construct the
binary tree.

Note:

You may assume that duplicates do not exist in the tree.

[Solution]

class Solution {
public:
// find element in an array
int find(vector<int> &array, int start, int end, int element){
for(int i = start; i <= end; ++i){
if(array[i] == element){
return i;
}
}
return -1;
}

// build tree
TreeNode *buildTree(vector<int> &preorder, int s1, int e1, vector<int> &inorder, int s2, int e2){
if(s1 > e1 || s2 > e2) return NULL;

// check invalid
int rootIndex = find(inorder, s2, e2, preorder[s1]);
if(-1 == rootIndex){
return NULL;
}

// set root node
TreeNode *root = new TreeNode(preorder[s1]);

// build left brunch and right brunch
int leftLen = rootIndex - s2;
TreeNode *left = buildTree(preorder, s1 + 1, s1 + leftLen, inorder, s2, rootIndex - 1);
TreeNode *right = buildTree(preorder, s1 + leftLen + 1, e1, inorder, rootIndex + 1, e2);

// set left brunch and right brunch
root->left = left;
root->right = right;

return root;
}

// build tree
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TreeNode *root = buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
return root;
}
};

说明：版权所有，转载请注明出处。Coder007的博客