[LintCode] 前序遍历和中序遍历树构造二叉树 Construct Binary Tree from Preorder and Inorder Traversal

根据前序遍历和中序遍历树构造二叉树.
注意事项
你可以假设树中不存在相同数值的节点.

样例
给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:
2
/ \
1 3

Given preorder and inorder traversal of a tree, construct the binary tree.
Notice
You may assume that duplicates do not exist in the tree.

Example
Given in-order [1,2,3] and pre-order [2,1,3], return a tree:
2
/ \
1 3

/**
 * 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
     */
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(null == preorder || null == inorder || preorder.length != inorder.length || preorder.length == 0) return null;
        return createTree(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
    }

    public TreeNode createTree(int[] preorder, int lp, int rp, int[] inorder, int li, int ri) {
        if(lp > rp || li > ri) return null;
        TreeNode root = new TreeNode(preorder[lp]);
        int inlo = location(inorder, preorder[lp]);
        root.left = createTree(preorder, lp+1, lp+inlo-li,inorder, li, inlo-1);
        root.right = createTree(preorder, lp+inlo-li+1, rp,inorder, inlo+1, ri);
        return root;
    }

    public int location(int[] nums, int key) {
        int i = 0;
        while(nums[i++] != key);
        return --i;
    }
}