03
04/2015
[LeetCode] Construct Binary Tree from Inorder and Postorder Traversal
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.
解题思路：
可以用递归的办法做。由后序遍历确定根节点，有中序遍历确定左右子树。每个子树有分别是一个后序遍历和一个中序遍历。代码如下。注意递归调用时位置的计算，后序遍历和中序遍历的长度一致。
/**
* Definition for binary tree
* 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, vector<int> &postorder) {
return getTreeNode(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size()-1);
}
TreeNode *getTreeNode(vector<int> &inorder, int startInOrder, int endInOrder,
vector<int> & postorder, int startPostOrder, int endPostOrder){
if(startInOrder == endInOrder){
return new TreeNode(postorder[startPostOrder]);
}else if(startInOrder > endInOrder){
return NULL;
}else{
int postNumber = postorder[endPostOrder];
//找到postNumber在中序遍历的位置
int inPosition=0;
for(int i=startInOrder; i<=endInOrder; i++){
if(inorder[i]==postNumber){
inPosition=i;
break;
}
}
TreeNode* node = new TreeNode(postNumber);
node->right = getTreeNode(inorder, inPosition+1,endInOrder, postorder, endPostOrder+inPosition-endInOrder, endPostOrder - 1);
node->left = getTreeNode(inorder, startInOrder, inPosition-1, postorder, startPostOrder, inPosition - 1 - startInOrder + startPostOrder);
return node;
}
}
};
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