【剑指offer】重建二叉树

题目描述

  输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

分析

1、二叉树的前序遍历序列一定是该树的根节点
2、中序遍历序列中根节点前面一定是该树的左子树，后面是该树的右子树

Java版本1

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
   public TreeNode reConstructBinaryTree(int [] preOrder,int [] inOrder) { 
		int preLength = preOrder.length;
		int inLength = inOrder.length;
		
		if(preLength == 0 && inLength == 0)
			return null;
		
		return reConstructBinaryTree(preOrder, inOrder, 0, preLength - 1, 0, inLength - 1);
	}
	
	private TreeNode reConstructBinaryTree(int[] preOrder, int[] inOrder, int pStart, int pEnd, int iStart, int iEnd) {

		TreeNode root = new TreeNode(preOrder[pStart]);
		if(pStart == pEnd && iStart == iEnd)
			return root;
		
		//在中序序列中找到根结点
		int rootIndex = 0;
		for(rootIndex = iStart; rootIndex < iEnd; rootIndex ++) {
			if(preOrder[pStart] == inOrder[rootIndex])
				break;
		}
		
		//划分左右子树
		int leftLength = rootIndex - iStart;
		int rightLength = iEnd - rootIndex;
		
		if(leftLength > 0)
			root.left = reConstructBinaryTree(preOrder, inOrder, pStart + 1, pStart + leftLength , iStart, rootIndex - 1 );
		
		if(rightLength > 0)
            root.right = reConstructBinaryTree(preOrder, inOrder, pStart + leftLength + 1, pEnd, rootIndex + 1, rootIndex + rightLength );
		return root;
	}
		
}

Java版本2

/**
*
 * Definition for binary tree
3
 * public class TreeNode {
4
 *     int val;
 
 *     TreeNode left;
6
 *     TreeNode right;
7
 *     TreeNode(int x) { val = x; }
8
 * }
9
 */

public class Solution {

   public TreeNode reConstructBinaryTree(int [] preOrder,int [] inOrder) { 

        int preLength = preOrder.length;

        int inLength = inOrder.length;

        if(preLength == 0 && inLength == 0)

            return null;

        return reConstructBinaryTree(preOrder, inOrder, 0, preLength - 1, 0, inLength - 1);

    }

    

    private TreeNode reConstructBinaryTree(int[] preOrder, int[] inOrder, int pStart, int pEnd, int iStart, int iEnd) {
        
       
        if(pStart > pEnd || iStart > iEnd)
            return null;
        
        //在中序序列中找到根结点
         TreeNode root = new TreeNode(preOrder[pStart]);
        for(int rootIndex = iStart; rootIndex <= iEnd; rootIndex ++) {

            if(preOrder[pStart] == inOrder[rootIndex]){
                //划分左右子树
              
                root.left = reConstructBinaryTree(preOrder, inOrder, pStart + 1, pStart + rootIndex  - iStart, iStart, rootIndex - 1 );

                root.right = reConstructBinaryTree(preOrder, inOrder,  rootIndex - iStart + pStart + 1, pEnd, rootIndex + 1, iEnd );

            }
        }
        return root;
    }
}

Python版本

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回构造的TreeNode根节点
    def reConstructBinaryTree(self, pre, tin):
        # write code here
        if len(pre) == 0:
            return None
        elif len(pre) == 1:
            return TreeNode(pre[0])
        else:
            root = TreeNode(pre[0])
            root.left = self.reConstructBinaryTree(pre[1:tin.index(pre[0]) + 1], tin[:tin.index(pre[0])])
            root.right = self.reConstructBinaryTree(pre[tin.index(pre[0]) + 1:], tin[tin.index(pre[0]) + 1:])
        return root