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根据中序遍历和后序遍历树构造二叉树
给出树的中序遍历： [1,2,3] 和后序遍历： [1,3,2]
返回如下的树：
2
/ \
1 3
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
def buildSubTree(inorder, inBeg, inEnd, postorder, postBeg, postEnd):
inLen = len(inorder)
postLen = len(postorder)
if inBeg == inEnd or inEnd > inLen or postBeg == postEnd or postEnd > postLen:
return None
root = TreeNode(postorder[postEnd - 1])
pos = inorder.index(root.val)
leftLen = pos - inBeg - 1
root.left = buildSubTree(inorder, inBeg, inBeg + leftLen + 1, postorder, postBeg, postBeg + leftLen + 1)
root.right = buildSubTree(inorder, inBeg + leftLen + 2, inEnd, postorder, postBeg + leftLen + 1, postEnd - 1)
return root
class Solution:
"""
@param inorder : A list of integers that inorder traversal of a tree
@param postorder : A list of integers that postorder traversal of a tree
@return : Root of a tree
"""
def buildTree(self, inorder, postorder):
# write your code here
return buildSubTree(inorder, 0, len(inorder), postorder, 0, len(postorder))
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
/**
*@param inorder : A list of integers that inorder traversal of a tree
*@param postorder : A list of integers that postorder traversal of a tree
*@return : Root of a tree
*/
public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
// write your code here
return buildSubTree(inorder.begin(),inorder.end(),postorder.begin(),postorder.end());
}
TreeNode *buildSubTree(vector<int>::iterator inBeg, vector<int>::iterator inEnd,
vector<int>::iterator postBeg, vector<int>::iterator postEnd)
{
if(inBeg == inEnd || postBeg == postEnd)
{
return NULL;
}//if
TreeNode *root = new TreeNode(*(postEnd - 1));
vector<int>::iterator pos = find(inBeg, inEnd, root->val);
if(pos != inEnd)
{
int leftLen = pos - inBeg - 1;
root->left = buildSubTree(inBeg,inBeg+leftLen+1, postBeg,postBeg+leftLen+1);
root->right = buildSubTree(inBeg+leftLen+2,inEnd, postBeg+leftLen+1, postEnd -1 );
}//if
return root;
}
};
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原文地址：http://blog.csdn.net/fly_yr/article/details/51685182
