RebuildTree

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题目描述[原题链接][https://www.acwing.com/problem/content/23/]

输入一棵二叉树前序遍历和中序遍历的结果，请重建该二叉树。

注意:

  • 二叉树中每个节点的值都互不相同；
  • 输入的前序遍历和中序遍历一定合法；

样例

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给定：
前序遍历是：[3, 9, 20, 15, 7]
中序遍历是：[9, 3, 15, 20, 7]

返回：[3, 9, 20, null, null, 15, 7, null, null, null, null]
返回的二叉树如下所示：
    3
   / \
  9  20
    /  \
   15   7

算法描述

通过先序遍历，找到每棵子树的根节点，在中序中找到其左右子树，递归建树，直到遍历完遍历序列；

C++代码

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    unordered_map<int,int> pos;
    
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        for(int i=0;i<n;i++){
            pos[inorder[i]] = i;
            
        }
        return dfs(preorder,inorder,0,n-1,0,n-1);
    }
    
    TreeNode* dfs(vector<int>&pre,vector<int>&in,int pl,int pr,int il,int ir){
        if(pl>pr)return NULL;
        int k = pos[pre[pl]]-il;
        TreeNode* root = new TreeNode(pre[pl]);
        root->left = dfs(pre,in,pl+1,pl+k,il,il+k-1);
        root->right = dfs(pre,in,pl+k+1,pr,il+k+1,ir);
        return root;
    }
};

Java代码

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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    HashMap<Integer,Integer> hm = new HashMap<>();
    int[] pre;
    int[] in;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        pre = new int[preorder.length];
        in = new int[inorder.length];
        System.arraycopy(preorder,0,pre,0,preorder.length);
        System.arraycopy(inorder,0,in,0,inorder.length);
        for(int i = 0; i<inorder.length;i++)hm.put(inorder[i],i);
        return dfs(0,pre.length-1,0,in.length-1);
    }
    
    public TreeNode dfs(int pl,int pr,int il,int ir){
        if(pl>pr)return null;
        TreeNode root = new TreeNode(pre[pl]);
        int k = hm.get(root.val);
        TreeNode left = dfs(pl+1,pl+k-il,il,k-1);
        TreeNode right = dfs(pl+k-il+1,pr,k+1,ir);
        root.left = left;
        root.right = right;
        return root;
    }
}