題目

根據(jù)前序遍歷和中序遍歷樹(shù)構(gòu)造二叉樹(shù).

注意事項(xiàng)

你可以假設(shè)樹(shù)中不存在相同數(shù)值的節(jié)點(diǎn)

樣例
給出中序遍歷：[1,2,3]和前序遍歷：[2,1,3]. 返回如下的樹(shù):

2
/
1 3

代碼

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     *@param preorder : A list of integers that preorder traversal of a tree
     *@param inorder : A list of integers that inorder traversal of a tree
     *@return : Root of a tree
     */
    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);
    }
}