前序遍历：根-左-右
中序遍历：左-根-右
后序遍历：左-右-根
遍历前序序列，每个元素找到它在中序序列中对应的位置，该位置左边代表左子树，右边代表右子树，递归建树
以样例为例，元素1，在中序序列中下标为3，所以0~2为左子树，4~7为右子树。
元素2，在中序序列中下标为2，所以0~1为左子树，没有右子树。
递归版：
1 class Solution { 2 public: 3 4 int k; 5 map<int, int> a; 6 Solution() 7 { 8 k = 0; 9 } 10 11 void built(int l, int r, TreeNode* &cur, vector<int> &pre) 12 { 13 if(l > r) 14 return; 15 16 cur = new TreeNode(pre[k]); 17 int t = k++; 18 19 built(l, a[pre[t]] - 1, cur->left, pre); 20 21 built(a[pre[t]] + 1, r, cur->right, pre); 22 } 23 24 25 TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) { 26 int len = pre.size(); 27 for(int i = 0; i < len; i++) 28 a[vin[i]] = i; 29 30 TreeNode* cur = NULL; 31 built(0, len - 1, cur, pre); 32 return cur; 33 } 34 };
非递归版：
1 struct StackNode{ 2 TreeNode* cur; 3 int l, r; 4 int vis; 5 StackNode(TreeNode *cur, int l, int r, int vis) 6 { 7 this->cur = cur; 8 this->l = l; 9 this->r = r; 10 this->vis = vis; 11 } 12 }; 13 14 class Solution { 15 public: 16 17 int k; 18 map<int, int> a; 19 stack<StackNode* > s; 20 Solution() 21 { 22 k = 0; 23 } 24 25 TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) { 26 int len = pre.size(); 27 int pos = 0; 28 for(int i = 0; i < len; i++) 29 a[vin[i]] = i; 30 31 TreeNode *cur = NULL; 32 if(!len) 33 return cur; 34 cur = new TreeNode(pre[pos++]); 35 StackNode *node = new StackNode(cur, 0, len - 1, 0); 36 s.push(node); 37 while(!s.empty()) 38 { 39 node = s.top(); 40 cur = node->cur; 41 42 if(node->l < a[cur->val] && node->vis < 1) 43 { 44 s.pop(); 45 node->vis = 1; 46 s.push(node); 47 cur->left = new TreeNode(pre[pos++]); 48 node = new StackNode(cur->left, node->l, a[cur->val] - 1, 0); 49 s.push(node); 50 continue; 51 } 52 53 if(node->r > a[cur->val] && node->vis < 2) 54 { 55 s.pop(); 56 node->vis = 2; 57 s.push(node); 58 cur->right = new TreeNode(pre[pos++]); 59 node = new StackNode(cur->right, a[cur->val] + 1, node->r, 0); 60 s.push(node); 61 continue; 62 } 63 64 s.pop(); 65 } 66 return cur; 67 } 68 };