leetCode(18):Construct Binary Tree from Preorder and Inorder (Inorder and Postorder) Traversal
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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); }};
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