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

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

Return the following binary tree:

  3
 / \
9  20
  /  \
 15   7

Solution

Language: Java

和第106题基本一致

/**
 * 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) {
        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;
    }
}