CODE 28: 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.

public TreeNode buildTree(int[] preorder, int[] inorder) {
// Start typing your Java solution below
// DO NOT write main() function
if (null == inorder || null == preorder || inorder.length <= 0
|| preorder.length <= 0) {
return null;
}
return dfs(inorder, preorder, 0, inorder.length - 1, 0,
preorder.length - 1);
}

TreeNode dfs(int[] inorder, int[] preorder, int inorderStart,
int inorderEnd, int preorderStart, int preorderEnd) {
TreeNode node = new TreeNode(preorder[preorderStart]);
int nodeIndexInInorder = findElement(inorder, preorder[preorderStart]);
int leftLength = nodeIndexInInorder - inorderStart;
int rightLength = inorderEnd - nodeIndexInInorder;
if (leftLength > 0) {
TreeNode left = dfs(inorder, preorder, inorderStart,
nodeIndexInInorder - 1, preorderStart + 1, preorderStart
+ leftLength);
node.left = left;
}
if (rightLength > 0) {
TreeNode right = dfs(inorder, preorder, nodeIndexInInorder + 1,
inorderEnd, preorderEnd - rightLength + 1, preorderEnd);
node.right = right;
}
return node;
}

int findElement(int[] array, int element) {
for (int i = 0; i < array.length; i++) {
if (array[i] == element) {
return i;
}
}
return -1;
}