剑指offer:面试题6、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。

代码示例

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Offer6 {
    private List<Integer> res = new ArrayList<>();

    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};
        Offer6 testObj = new Offer6();
        TreeNode root = testObj.reBuildBinaryTree(pre, in);
        //用先序遍历验证重建的二叉树是否准确
        testObj.preOrder(root);
        System.out.println(testObj.res);
    }

    static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int val) {
            this.val = val;
        }
    }
    //建立中序遍历序列数值和索引的映射关系
    private Map<Integer, Integer> indexForInorders = new HashMap<>();

    //根据先序遍历序列和中序遍历序列重建二叉树
    public TreeNode reBuildBinaryTree(int[] pre, int[] in) {
        for (int i = 0; i < in.length; i++) {
            indexForInorders.put(in[i], i);
        }
        return reBuildBinaryTree(pre, 0, pre.length - 1, 0);
    }
    //先处理根节点，然后再递归处理左子树和右子树
    private TreeNode reBuildBinaryTree(int[] pre, int preLeft, int preRight, int inLeft) {
        if (preLeft > preRight) {
            return null;
        }
        TreeNode root = new TreeNode(pre[preLeft]);//先序遍历第一个结点为root结点
        int inIndex = indexForInorders.get(root.val);//找到中序遍历序列的root结点索引将其分割成左右子树
        int leftTreeSize = inIndex - inLeft;//左子树的个数
        root.left = reBuildBinaryTree(pre, preLeft + 1, preLeft + leftTreeSize, inLeft);
        root.right = reBuildBinaryTree(pre, preLeft + leftTreeSize + 1, preRight, inLeft + leftTreeSize + 1);
        return root;
    }
    //先序遍历二叉树，递归实现
    private void preOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        res.add(root.val);
        preOrder(root.left);
        preOrder(root.right);
    }
}