题解 | #从中序与后序遍历序列构造二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param inorder int整型vector 中序遍历序列
     * @param postorder int整型vector 后序遍历序列
     * @return TreeNode类
     */
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        // write code here
        int n= inorder.size(),m = postorder.size();
        return maketree(inorder,postorder,0,n-1,0,n-1);
    }
    TreeNode*maketree( vector<int> &id,vector<int> &pd,int l,int r,int j,int k) {
        if(l>r || j>k) return nullptr;
        if(l==r) 
            return new TreeNode(id[l]);
        TreeNode * x = new TreeNode(pd[k]);
        int m = l;
        for(int i=l;i<=r;i++) {
            if(id[i] == pd[k]) {
                m = i;
                break;
            }
        }
        int left = k - (r-m);
        x->left = maketree(id,pd,l,m-1,j,left-1);
        x->right = maketree(id,pd,m+1,r,left,k-1);
        return x;

    }
};