Construct Binary Tree from Preorder and Inorder Traversal - LeetCode

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本文介绍了一种通过前序遍历和中序遍历构建二叉树的方法。利用递归算法找到根节点，并根据中序遍历划分左右子树，递归构造整棵树。

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

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

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public class Solution {
    private int findPosition(int[] arr, int start, int end, int key) {
        int i;
        for (i = start; i <= end; i++) {
            if (arr[i] == key) {
                return i;
            }
        }
        return -1;
    }
    
    private TreeNode myBuildTree(int[] inorder, int instart, int inend,
                                 int[] preorder, int prestart, int preend) {
        if (instart > inend) {
            return null;
        }
        
        TreeNode root = new TreeNode(preorder[prestart]);
        int position = findPosition(inorder, instart, inend, preorder[prestart]);
        
        root.left = myBuildTree(inorder, instart, position - 1,
                                preorder, prestart + 1, prestart + position - instart);
        root.right = myBuildTree(inorder, position + 1, inend,
                                 preorder, position - inend + preend + 1, preend);
        return root;
    }
    
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (inorder.length != preorder.length) {
            return null;
        }
        return myBuildTree(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
    }
}