Construct Binary Tree from Preorder and Inorder Traversal – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given two integer arrays preorder
and inorder
where preorder
is the preorder traversal of a binary tree and inorder
is the inorder traversal of the same tree, construct and return the binary tree.
Example 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]
Constraints:
1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder
andinorder
consist of unique values.- Each value of
inorder
also appears inpreorder
. preorder
is guaranteed to be the preorder traversal of the tree.inorder
is guaranteed to be the inorder traversal of the tree.
C++ Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int rootIdx = 0;
return build(preorder, inorder, rootIdx, 0, inorder.size()-1);
}
TreeNode* build(vector<int>& preorder, vector<int>& inorder, int& rootIdx, int left, int right) {
if (left > right) return NULL;
int pivot = left; // find the root from inorder
while(inorder[pivot] != preorder[rootIdx]) pivot++;
rootIdx++;
TreeNode* newNode = new TreeNode(inorder[pivot]);
newNode->left = build(preorder, inorder, rootIdx, left, pivot-1);
newNode->right = build(preorder, inorder, rootIdx, pivot+1, right);
return newNode;
}
};
Java Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution
public TreeNode buildTree(int[] preorder, int[] inorder) {
return helper(0, 0, inorder.length - 1, preorder, inorder);
}
public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
if (preStart > preorder.length - 1 || inStart > inEnd) {
return null;
}
TreeNode root = new TreeNode(preorder[preStart]);
int inIndex = 0; // Index of current root in inorder
for (int i = inStart; i <= inEnd; i++) {
if (inorder[i] == root.val) {
inIndex = i;
}
}
root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
return root;
}
Python 3 Construct Binary Tree from Preorder and Inorder Traversal LeetCode Solution
def buildTree(self, preorder, inorder):
if inorder:
ind = inorder.index(preorder.pop(0))
root = TreeNode(inorder[ind])
root.left = self.buildTree(preorder, inorder[0:ind])
root.right = self.buildTree(preorder, inorder[ind+1:])
return root
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