leetcode 105. construct binary tree from preorder and inorder traversal

105. Construct Binary Tree from Preorder and Inorder Traversal

Difficulty:: 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

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2
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

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2
3
4
5
  3
/
9 20
/
15 7

Solution

Language: Java

和第106题基本一致

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* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (inorder == null || preorder == null || inorder.length == 0 || preorder.length == 0) {
return null;
}
int len = preorder.length;
Map<Integer, Integer> indexMap = new HashMap<>();
for (int i = 0; i < len; i++) {
indexMap.put(inorder[i], i);
}
return helper(inorder, preorder, 0, len - 1, 0, len - 1, indexMap);
}

private TreeNode helper(int[] inorder, int[] preorder, int iStart, int iEnd, int pStart, int pEnd, Map<Integer, Integer> map) {
if (iStart > iEnd || pStart > pEnd) {
return null;
}
int pivot = preorder[pStart];
TreeNode root = new TreeNode(pivot);
int pivotIndex = map.get(pivot);
int leftLen = pivotIndex - iStart;
root.left = helper(inorder, preorder, iStart, pivotIndex - 1, pStart + 1, pStart + leftLen, map);
root.right = helper(inorder, preorder, pivotIndex + 1, iEnd, pStart + leftLen + 1, pEnd, map);
return root;
}
}