leetCode(18):Construct Binary Tree from Preorder and Inorder

Given preorder and inorder (Inorder and Postorder) traversal of a tree, construct the binary tree.

Note:You may assume that duplicates do not exist in the tree.

前序和后序的特点是根结点要么在最前面、要么在最后面。已知根结点后，，以根结点为界将中序遍历的结果分成两段，然后递归即可还原二叉树。有两点需要注意：首先二叉树中不能有重复的结点；其次，已知前序和后序遍历结果是无法还原二叉树的。

已知前序和中序，构建二叉树：

/** * 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>& preorder,int from1,int to1,vector<int>& inorder,int from2,int to2){if(from1>to1)return NULL;int i=from2;for(;i<=to2;++i){if(inorder[i]==preorder[from1])break;}TreeNode* root=new TreeNode(inorder[i]);root->left=buildTree(preorder,from1+1,from1+i-from2,inorder,from2,i-1);root->right=buildTree(preorder,from1+i-from2+1,to1,inorder,i+1,to2);return root;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {return buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);}};已知中序和后序，构建二叉树：

/** * 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,int from1,int to1,vector<int>& postorder,int from2,int to2){if(from1>to1)return NULL;// if(from1==to1)// {// TreeNode* tmp=new TreeNode(inorder[from1]);// return tmp;// }int i=from1;for(;i<=to1;++i){if(inorder[i]==postorder[to2])break;}TreeNode* root=new TreeNode(inorder[i]);root->left=buildTree(inorder,from1,i-1,postorder,from2,from2+i-from1-1);root->right=buildTree(inorder,i+1,to1,postorder,from2+i-from1,to2-1);return root;}TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {return buildTree(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);}};

只能昏昏沉沉地沿着青草和泥土的气息前进。