leetcode第一刷_Construct Binary Tree from Inorder and Postorder Traversal

这道题是为数不多的感觉在读本科的时候见过的问题。

人工构造的过程是如何呢。兴许遍历最后一个节点一定是整棵树的根节点。从中序遍历中查找到这个元素，就能够把树分为两颗子树，这个元素左側的递归构造左子树，右側的递归构造右子树。元素本身分配空间，作为根节点。

于set和map容器不同的是。vector容器不含find的成员函数。应该用stl的库函数，好在返回的也是迭代器，而vector的迭代器之间是能够做减法的。偏移量非常方便的得到。

TreeNode *buildRec(vector<int> &inorder, int si, vector<int> &postorder, int so, int len){if(len <= 0)    return NULL;TreeNode *root = new TreeNode(postorder[so]);int index = find(inorder.begin(), inorder.end(), postorder[so]) - inorder.begin();int newlen = index - si;root->left = buildRec(inorder, si, postorder, so-len+newlen, newlen);root->right = buildRec(inorder, index+1, postorder, so-1, len-newlen-1);return root;
}class Solution {
public:TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {return buildRec(inorder, 0, postorder, inorder.size()-1, inorder.size());}
};

转载于:https://www.cnblogs.com/blfshiye/p/5135053.html

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