Skip to content

105. Construct Binary Tree from Preorder and Inorder Traversal

🔗 Source

Intuition

Approach

Complexity

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(n)\)

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer, Integer> map = new HashMap<>();
        int end = preorder.length - 1;
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return build(preorder, 0, end, inorder, 0, end, map);
    }

    private TreeNode build(int[] preorder, int preStart, int preEnd, 
                           int[]inorder, int inStart, int inEnd, 
                           Map<Integer,Integer> map) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inRoot = map.get(root.val);
        int numsLeft = inRoot - inStart;
        root.left = build(preorder, preStart + 1, preStart + numsLeft, 
                          inorder, inStart, inRoot - 1, map);
        root.right = build(preorder, preStart + numsLeft + 1, preEnd, 
                           inorder, inRoot + 1, inEnd, map);
        return root;
    }

}