由前/中/后遍历序列构建二叉树

看一下前序和中序有什么特点，前序1,2,4,7,3,5,6,8 ，中序4,7,2,1,5,3,8,6；

当前要构造的树（可能是整体的某个子树）所需的前序/中序遍历序列肯定是整体所求树的前序/中序遍历序列的某个子序列，我们用 preLeft, preRight, inLeft, inRight 分别表示该子序列在整体前序/中序序列中的起始位置。

1. 前序中左起第一位1肯定是根结点 root，我们可以根据其值 rootVal 找到中序中根结点的位置记为 mid；
2. 中序中根结点左边就是左子树结点，右边就是右子树结点。即[左子树结点，根结点，右子树结点]，我们就可以得出左子树结点个数为 mid-inLeft；
3. 前序中结点分布应该是：[根结点，左子树结点，右子树结点]；
4. 根据前一步确定的左子树个数，可以确定前序中左子树结点和右子树结点的范围；
5. 如果我们要前序遍历生成二叉树的话，下一层递归应该是：
   • 左子树：root->left = helper(preorder, inorder, 左子树前序起始点, 左子树前序终止点, 左子树中序起始点, 左子树中序终止点);
   • 右子树：root->right = helper(preorder, inorder, 右子树前序起始点, 右子树前序终止点, 右子树中序起始点, 右子树中序终止点);
6. 每一层递归都要返回当前根结点root；

全部代码如下：

class Solution {
private:TreeNode* helper(vector<int>& preorder, vector<int>& inorder, int preLeft, int preRight, int inLeft, int inRight) {if (inLeft >= inRight) return nullptr;int rootVal = preorder[preLeft];int mid = inLeft;for (; mid<inRight; ++mid) if (inorder[mid] == rootVal) break;TreeNode* root = new TreeNode(rootVal);root->left = helper(preorder, inorder, preLeft+1, preLeft+1+mid-inLeft, inLeft, mid);root->right = helper(preorder, inorder, preLeft+1+mid-inLeft, preRight, mid+1, inRight);return root;}
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {return helper(preorder, inorder, 0, preorder.size(), 0, inorder.size());}
};

与前序中序构造二叉树类似：

class Solution {
private:TreeNode* helper(vector<int>& inorder, vector<int>& postorder, int inLeft, int inRight, int postLeft, int postRight) {if (inLeft >= inRight) return nullptr;int rootVal = postorder[postRight-1];int mid = inLeft;for (; mid<inRight; ++mid) if (inorder[mid] == rootVal) break;TreeNode* root = new TreeNode(rootVal);root->left = helper(inorder, postorder, inLeft, mid, postLeft, postLeft+mid-inLeft);root->right = helper(inorder, postorder, mid+1, inRight, postLeft+mid-inLeft, postRight-1);return root;}
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {return helper(inorder, postorder, 0, inorder.size(), 0, postorder.size());}
};

注意，我们之前提到前序/后序序列并不能唯一确定一棵二叉树（实际上，只有每个节点度为2或者0的时候前序和后序才能唯一确定一颗二叉树），所以本题题目说明中有：

• 每个输入保证至少有一个答案。如果有多个答案，可以返回其中一个。

class Solution {
private:TreeNode* helper(vector<int>& preorder, vector<int>& postorder, int preLeft, int preRight, int postLeft, int postRight) {if (preLeft > preRight) return nullptr;int rootVal = preorder[preLeft];TreeNode* root = new TreeNode(rootVal);if (preLeft == preRight) return root;int mid = 0;for (; mid<postRight; ++mid) if (postorder[mid] == preorder[preLeft+1]) break;root->left = helper(preorder, postorder, preLeft+1, preLeft+1+mid-postLeft, postLeft, mid);root->right = helper(preorder, postorder, preLeft+2+mid-postLeft, preRight, mid+1, postRight-1);return root;}
public:TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder) {return helper(preorder, postorder, 0, preorder.size()-1, 0, postorder.size()-1);}
};