Construct Binary Tree from Preorder and Inorder Traversal

Desicription

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

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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
TreeNode* generateTree(int preStart, int inStart, int inEnd, vector<int>& preorder, vector<int>& inorder) {
if(preStart > preorder.size()-1 || inStart > inEnd)
return NULL;
TreeNode* root = new TreeNode(preorder[preStart]);
int inIndex = 0;
for(int i = inStart; i <= inEnd; i++) {
if(inorder[i] == preorder[preStart]){
inIndex = i;
break;
}
}
root->left = generateTree(preStart+1, inStart, inIndex-1, preorder, inorder);
root->right = generateTree(preStart+inIndex-inStart+1, inIndex+1, inEnd, preorder, inorder);
return root;
}
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return generateTree(0, 0, inorder.size()-1, preorder, inorder);
}
};