105. Construct Binary Tree from Preorder and Inorder Traversal
Similar question:
Time = O(n), Space = O(n)
/**
* 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) {
Map<Integer, Integer> inMap = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
inMap.put(inorder[i], i);
}
return buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);
}
public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, Map<Integer, Integer> inMap) {
if (preStart > preEnd || inStart > inEnd) {
return null;
}
TreeNode root = new TreeNode(preorder[preStart]);
int inRoot = inMap.get(root.val);
int numsLeft = inRoot - inStart;
root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);
root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);
return root;
}
}