题7：重建二叉树(Java)

我GitHub上的：剑指offer题解​​​​​​​

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。

思路：前序遍历的第一个值为根节点的值，使用这个值将中序遍历结果分成两部分，左部分为树的左子树中序遍历结果，右部分为树的右子树中序遍历的结果

 

二刷：

package shuti;

import test.TreeNode;

public class ReConstructBinaryTree {

	/**
	 * 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。
	 * @param pre
	 * @param in
	 * @return
	 */

	public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
		 TreeNode root = reConstructBinaryTree(pre, 0,pre.length - 1, in, 0, in.length - 1);
			return root;
	        
	}
	
	private static TreeNode reConstructBinaryTree(int[] pre, int startPre, int endPre, int[] in, int startIn, int endIn) {
		if(startPre > endPre || startIn > endIn) {
			return null;
		}
		TreeNode root = new TreeNode(pre[startPre]);
		for(int i = startIn; i <= endIn; i++) {
			if(pre[startPre] == in[i]) {
				root.left = reConstructBinaryTree(pre, startPre+1, startPre + i - startIn, in, startIn, i - 1);
				root.right = reConstructBinaryTree(pre, startPre + i - startIn + 1, endPre, in, i + 1, endIn );
				break;
			}
		}
		
		return root;
		
	}
		
	public static void main(String[] args) {
		int[] pre = {1,2,4,7,3,5,6,8};
		int[] in = {4,7,2,1,5,3,8,6};
		reConstructBinaryTree(pre, in);
	}
}

一刷时 

import java.util.HashMap;
import java.util.Map;

public class ReConstructBinaryTree {
	//用来缓存中序遍历数组的每个值对应的索引
	private static Map inOrderNumsIdx = new HashMap<>();
	
	public static TreeNode reConstructBinaryTree(int[] pre,int[] in) {
		for(int i = 0; i < in.length; i++){
			inOrderNumsIdx.put(in[i], i);
		}
		return reConstructBinaryTree(pre, 0, pre.length - 1, in, 0, in.length - 1);
		
	}
	
	public static TreeNode reConstructBinaryTree(int[] pre, int preL, int preR,int[] in,int inL, int inR) {
		if (preL == preR) {
			return new TreeNode(pre[preL]);
		}
		if (preL > preR || inL > inR) {
			return null;
		}
		TreeNode root = new TreeNode(pre[preL]);
		int inIdx = inOrderNumsIdx.get(root.val);//inOrder中根节点对应的索引
		int leftTreeSize = inIdx - inL;
		root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, in, inL, inL + leftTreeSize - 1);
		root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, in, inL + leftTreeSize + 1, inR);
		return root;
		
	}
	public static void main(String[] args) {
		int[] pre = {1,2,4,7,3,5,6,8};
		int[] in = {4,7,2,1,5,3,8,6};
		TreeNode tree = new TreeNode(0);
		tree = reConstructBinaryTree(pre, in);
	}
}

 

public class TreeNode {
	 int val;
	 TreeNode left;
	 TreeNode right;
	 TreeNode(int x) { 
		 val = x; 
	 }
}