LeetCode Java Solutions

105. Construct Binary Tree from Preorder and Inorder Traversal

Medium

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

3
/ \
9 20
/ \
15 7

Solution

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return helper(0, 0, inorder.length - 1, preorder, inorder);
}
private TreeNode helper(int preStart, int inStart, int inEnd, int[] preOrder, int[] inOrder) {
//Boundary Conditions.
if(preStart > preOrder.length-1) {
return null;
}
if(inStart > inEnd) {
return null;
}
TreeNode root = new TreeNode(preOrder[preStart]);
int inIndex = 0;
for(int i=inStart; i<= inEnd; i++) {
if(root.val == inOrder[i]) {
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;
}
}