Construct Binary Tree from Preorder and Inorder Traversal

描述

Given preorder and inorder traversal of a tree, construct the binary tree.

Note: You may assume that duplicates do not exist in the tree.

分析

无

代码

// Construct Binary Tree from Preorder and Inorder Traversal
// 递归，时间复杂度O(n)，空间复杂度O(\logn)
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return buildTree(preorder, 0, preorder.length,
                inorder, 0, inorder.length);
    }

    TreeNode buildTree(int[] preorder, int begin1, int end1,
                       int[] inorder, int begin2, int end2) {
        if (begin1 == end1) return null;
        if (begin2 == end2) return null;

        TreeNode root = new TreeNode(preorder[begin1]);

        int inRootPos = find(inorder, begin2, end2, preorder[begin1]);
        int leftSize = inRootPos - begin2;

        root.left = buildTree(preorder, begin1 + 1, begin1 + leftSize + 1, 
                inorder, begin2, begin2 + leftSize);
        root.right = buildTree(preorder, begin1 + leftSize + 1, end1,
                inorder, inRootPos + 1, end2);

        return root;
    }
    private static int find(int[] array, int begin, int end, int val) {
        for (int i = begin; i < end; ++i) {
            if (array[i] == val) return i;
        }
        return -1;
    }
}

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