C++详解Leetcode:106. Construct Binary Tree from Inorder and Postorder Traversal

原题

思路

通过二叉树的中序遍历和后序遍历来构建二叉树，通过递归可以很简单的解决

code

/** * 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:    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        int l1 = 0;        int l2 = 0;        int r1 = inorder.size() - 1;        int r2 = postorder.size() - 1;        TreeNode *s = (TreeNode*)malloc(sizeof(TreeNode));        //s = NULL;        s = CreateBT(inorder, postorder, l1, r1, l2, r2);        return s;    }    TreeNode* CreateBT(vector<int>& inorder, vector<int>& postorder, int l1, int r1, int l2, int r2)    {        TreeNode* s;        int i;        if (l1 > r1)            return NULL;        s = (TreeNode*)malloc(sizeof(TreeNode));        s->left = s->right = NULL;        for (i = l1; i <= r1; i++)        {            if (inorder[i] == postorder[r2])            {                break;            }        }        s->val = inorder[i];        s->left = CreateBT(inorder, postorder, l1, i - 1, l2, l2 + i -l1 - 1);        s->right = CreateBT(inorder, postorder, i + 1, r1, l2 + i - l1, r2 - 1);        return s;    }};