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

问题

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

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ExampleGiven in-order [1,2,3]
and pre-order [2,1,3]
, return a tree:

分析

以这个二叉树为例，前序是1245367，中序是4251637。我们知道前序的第一个一定是根结点。所以先把1找出来，然后在中序里边找到1，那么1左边的一定是1的左子树，1右边的一定是1的右子树，然后递归即可。里边需要使用到前序，中序和后序的知识，大家自行查阅资料学习。

代码

/**
 * 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) {
        // write your code here
        return buildTree(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    private TreeNode buildTree(int[] preorder,int pleft,int pright, int[] inorder,int ileft,int iright){
        if(pleft>pright||ileft>iright){
            return null;
        }
        int i=preorder[pleft];
        TreeNode node=new TreeNode(i);
        int index=ileft;
        while(inorder[index]!=i){
            index++;
        }
        int leftLength=index-ileft;
        node.left=buildTree(preorder,pleft+1,pleft+leftLength,inorder,ileft,index-1);
        node.right=buildTree(preorder,pleft+leftLength+1,pright,inorder,index+1,iright);
        return node;
    }
}