1. 题目

类似题目LeetCode 105. 已知前序&中序 求二叉树

2. 解题

2.1 递归

递归法，后序最后一个是根节点

class Solution {
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {unordered_map<int,int> m;//哈希表for(int i = 0; i < inorder.size(); ++i) {m[inorder[i]] = i;//方便后面查找位置}return build(postorder, inorder,0,postorder.size()-1,0,inorder.size()-1,m);}TreeNode* build(vector<int>& postorder, vector<int>& inorder, int pS, int pE, int iS, int iE, unordered_map<int,int> &m){if(pS > pE)return NULL;TreeNode *root = new TreeNode(postorder[pE]);int leftlen = m[postorder[pE]]-iS;root->left = build(postorder,inorder,pS,pS+leftlen-1,iS,m[postorder[pE]]-1,m);root->right = build(postorder,inorder,pS+leftlen,pE-1,m[postorder[pE]]+1,iE,m);return root;}
};

2.2 循环

class Solution {    //别人写的 循环
public:TreeNode* buildTree(vector<int>& in, vector<int>& post) {if (in.empty()) return NULL;stack<TreeNode*> S;TreeNode *root = new TreeNode(post.back()), *cur = root;S.push(root);for (int i = post.size() - 2, j = in.size() - 1; i >= 0; i--) {TreeNode *back = NULL, *cur = new TreeNode(post[i]);while (!S.empty() && S.top()->val == in[j]) {back = S.top(), S.pop(), j--;}if (back)back->left = cur;elseS.top()->right = cur;S.push(cur);}return root;}
};