leetcode -day23 Construct Binary Tree from Inorder and Postorder Traversal & Construct Binary Tree f

1、

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

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

代码：

class Solution {
public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
TreeNode* root = NULL;
int len1 = inorder.size();
int len2 = postorder.size();
if(len1<1 || len2<1 || len1!=len2){
return root;
}
return buildTreeCore(inorder,0,len1-1,postorder,0,len2-1);
}
TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&postorder, int startIndex2, int endIndex2){
if(endIndex1 < startIndex1  ){
return NULL;
}
TreeNode* root = new TreeNode(postorder[endIndex2]);
int index = 0;
for(int i=startIndex1; i<=endIndex1; ++i){
if(inorder[i] == postorder[endIndex2]){
index = i;
break;
}
}
int leftLen = index-startIndex1;
TreeNode* leftNode = NULL;
TreeNode* rightNode = NULL;

leftNode = buildTreeCore(inorder,startIndex1,index-1,postorder,startIndex2,startIndex2+leftLen-1);
rightNode = buildTreeCore(inorder,index+1,endIndex1,postorder,startIndex2+leftLen,endIndex2-1);

root->left = leftNode;
root->right = rightNode;
return root;
}
};

2、Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

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

class Solution {
public:
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
TreeNode* root = NULL;
int len1 = inorder.size();
int len2 = preorder.size();
if(len1<1 || len2<1 || len1!=len2){
return root;
}
return buildTreeCore(inorder,0,len1-1,preorder,0,len2-1);
}
TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&preorder, int startIndex2, int endIndex2){
if(endIndex1 < startIndex1  ){
return NULL;
}
TreeNode* root = new TreeNode(preorder[startIndex2]);
int index = 0;
for(int i=startIndex1; i<=endIndex1; ++i){
if(inorder[i] == preorder[startIndex2]){
index = i;
break;
}
}
int leftLen = index-startIndex1;
TreeNode* leftNode = NULL;
TreeNode* rightNode = NULL;

leftNode = buildTreeCore(inorder,startIndex1,index-1,preorder,startIndex2+1,startIndex2+leftLen);
rightNode = buildTreeCore(inorder,index+1,endIndex1,preorder,startIndex2+leftLen+1,endIndex2);

root->left = leftNode;
root->right = rightNode;
return root;
}
};