2020-09-25 刷题记录

106. 从中序与后序遍历序列构造二叉树

思路：

根据后序遍历的最后一个元素是根的性质逐步构造出二叉树。

代码：

/**
 * 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:

    vector<int> a, b;
    
   TreeNode * dfs(int al, int ae, int bl, int be){
       if(al > ae) return NULL;
       TreeNode * res = new TreeNode(b[be]);
       int idx;
       for(int i = al; i <= ae; i ++) if(a[i] == b[be]) { idx = i; break; }
       res->left = dfs(al, idx - 1, bl, bl + idx - al - 1);
       res->right = dfs(idx + 1, ae,bl + idx - al, be - 1);
       return res;
    }
    
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        a = inorder;
        b = postorder;
        TreeNode * res = dfs(0, a.size() - 1, 0, b.size() - 1);
        return res;
    }
};