重建二叉树

2020-12-15

剑指 Offer 07. 重建二叉树

难度中等423

输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。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，给出

1 2	前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

限制：

1	0 <= 节点个数 <= 5000

注意：本题与主站 105 题重复：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

递归法

每次递归中从前序遍历中选择第一个元素作为根节点，递归地生成左子树和右子树。在前序遍历中要想获取左子树的节点数目和右子树的节点的数目，需要借助中序遍历的根的index计算，计算完成后，即可正确获得左子树的根的位置和右子树的根的位置，从而可以递归的进行生成其左右子树。
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递归写法

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    
    Map<Integer,Integer> indexMap ;
    int []pre;
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        this.pre = pre;
        indexMap = new HashMap<>();
        for(int i=0;i<in.length;i++){
            indexMap.put(in[i],i);
        }
        
        TreeNode root = build(0,pre.length-1,0,in.length-1);
        return root;
    }
    
    public TreeNode build(int pStart,int pEnd,int inStart,int inEnd){
        if(pStart>pEnd)return null;
        TreeNode root = new TreeNode(pre[pStart]);
        if(pStart==pEnd)return root;
        
        int rootIndex = indexMap.get(pre[pStart]);
        int leftNum = rootIndex - inStart;
        int rightNum = inEnd - rootIndex;
        // 构建左子树
        TreeNode left = build(pStart+1, pStart + leftNum, inStart, rootIndex - 1);
        // 构建右子树
        TreeNode right = build(pEnd - rightNum + 1, pEnd, rootIndex+1, inEnd);
        root.left = left;
        root.right = right;
        return root;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
        int length = preorder.length;
        for (int i = 0; i < length; i++) {//将值与下标进行一一映射，加快查找过程
            indexMap.put(inorder[i], i);
        }
        TreeNode root = buildTree(preorder, 0, length - 1, inorder, 0, length - 1, indexMap);
        return root;
    }

    public TreeNode buildTree(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd, Map<Integer, Integer> indexMap) {
        if (preorderStart > preorderEnd) {
            return null;
        }
        int rootVal = preorder[preorderStart];//前序遍历的第一个节点是该子树的根节点
        TreeNode root = new TreeNode(rootVal);
        if (preorderStart == preorderEnd) {
            return root;
        } else {
            int rootIndex = indexMap.get(rootVal);//获取中序遍历中根节点的下标
            int leftNodesNumber = rootIndex - inorderStart;//计算左子树的节点的数量
            int rightNodesNumber = inorderEnd - rootIndex;//计算右子树节点的数量
            TreeNode leftSubtree = buildTree(preorder, preorderStart + 1, preorderStart + leftNodesNumber, inorder, inorderStart, rootIndex - 1, indexMap);//左子树的preorder开始位置是preorderStart + 1，结束位置是preorderStart + leftNodesNumber,inorder的开始位置是inorderStart，结束位置是rootIndex-1
            TreeNode rightSubtree = buildTree(preorder, preorderEnd - rightNodesNumber + 1, preorderEnd, inorder, rootIndex + 1, inorderEnd, indexMap);
            root.left = leftSubtree;
            root.right = rightSubtree;
            return root;
        }
    }
}

参考文献

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/solution/mian-shi-ti-07-zhong-jian-er-cha-shu-by-leetcode-s/
来源：力扣（LeetCode）
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