【leetcode刷题笔记】Construct Binary Tree from Preorder and Inorder Traversal

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

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

---

类似http://www.cnblogs.com/sunshineatnoon/p/3854935.html

只是子树的前序和中序遍历序列分别更新为：

//左子树：
left_prestart = prestart+1
left_preend = prestart+index-instart
//右子树
right_prestart = prestart+index-instart+1
right_preend =  preend

代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int InorderIndex(int[] inorder,int key){
        if(inorder == null || inorder.length == 0)
            return -1;
        
        for(int i = 0;i < inorder.length;i++)
            if(inorder[i] == key)
                return i;
        
        return -1;
    }
    public TreeNode buildTreeRec(int[] preoder,int[] inorder,int prestart,int preend,int instart,int inend){
        if(instart > inend)
            return null;
        TreeNode root = new TreeNode(preoder[prestart]);
        int index = InorderIndex(inorder, root.val);
        root.left = buildTreeRec(preoder, inorder, prestart+1, prestart+index-instart, instart, index-1);
        root.right = buildTreeRec(preoder, inorder, prestart+index-instart+1, preend, index+1, inend);
        
        return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return buildTreeRec(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
    }
}