【LeetCode】Construct Binary Tree from Inorder and Postorder Traversal

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题解：就是用中序和后序重建二叉树保证每个点不重复值上

这里顺便列下中序和先序的代码，思路都一样用map存储可以节省查找时间，注意边界处理

class Solution {
private:
        unordered_map<int, int> inm; // inorder map [inorder[i], i]

public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int n = inorder.size(), i = 0;
        for(auto val: inorder) inm[val] = i++; // build inm for dfs 
        
        return dfs(inorder, 0, n - 1, postorder, 0, n - 1);
    }
    
    TreeNode* dfs(vector<int>& inorder, int ileft, int iright, vector<int>& postorder, int pleft, int pright) {
        if(ileft > iright) return nullptr;
        
        int val = postorder[pright]; // root value
        TreeNode *root = new TreeNode(val);
        if(ileft == iright) return root;
        
        int iroot = inm[val];
        int nleft = iroot - ileft; // length of left subtree
        root->right = dfs(inorder, iroot + 1, iright, postorder, pleft + nleft, pright - 1);
        root->left = dfs(inorder, ileft, iroot - 1, postorder, pleft, pleft + nleft - 1);
        
        return root;
    }
};

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    unordered_map<int,int> mp;    
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        int i=0,n=inorder.size();
        for(auto x:inorder) mp[x]=i++;
        return dfs(preorder,0,n-1,inorder,0,n-1);
    }
    TreeNode *dfs(vector<int> pre,int pL,int pR,vector<int> in,int inL,int inR)
    {
        if(inL>inR) return nullptr;
        int val=pre[pL];
        TreeNode *root=new TreeNode(val);
        if(inL==inR) return root;
        int dis=mp[val]-inL;
        root->left=dfs(pre,pL+1,pL+dis,in,inL,inL+dis-1);
        root->right=dfs(pre,pL+dis+1,pR,in,inL+dis+1,inR);
        return root;
    }
};