[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
Java:
public TreeNode buildTreePostIn(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null || inorder.length != postorder.length)
return null;
HashMap<Integer, Integer> hm = new HashMap<Integer,Integer>();
for (int i=0;i<inorder.length;++i)
hm.put(inorder[i], i);
return buildTreePostIn(inorder, 0, inorder.length-1, postorder, 0,
postorder.length-1,hm);
}
private TreeNode buildTreePostIn(int[] inorder, int is, int ie, int[] postorder, int ps, int pe,
HashMap<Integer,Integer> hm){
if (ps>pe || is>ie) return null;
TreeNode root = new TreeNode(postorder[pe]);
int ri = hm.get(postorder[pe]);
TreeNode leftchild = buildTreePostIn(inorder, is, ri-1, postorder, ps, ps+ri-is-1, hm);
TreeNode rightchild = buildTreePostIn(inorder,ri+1, ie, postorder, ps+ri-is, pe-1, hm);
root.left = leftchild;
root.right = rightchild;
return root;
}
Python:
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# @param inorder, a list of integers
# @param postorder, a list of integers
# @return a tree node
def buildTree(self, inorder, postorder):
lookup = {}
for i, num in enumerate(inorder):
lookup[num] = i
return self.buildTreeRecu(lookup, postorder, inorder, len(postorder), 0, len(inorder))
def buildTreeRecu(self, lookup, postorder, inorder, post_end, in_start, in_end):
if in_start == in_end:
return None
node = TreeNode(postorder[post_end - 1])
i = lookup[postorder[post_end - 1]]
node.left = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1 - (in_end - i - 1), in_start, i)
node.right = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1, i + 1, in_end)
return node
if __name__ == "__main__":
inorder = [2, 1, 3]
postorder = [2, 3, 1]
result = Solution().buildTree(inorder, postorder)
print(result.val)
print(result.left.val)
print(result.right.val)
C++:
// Time: O(n)
// Space: O(n)
/**
* 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, vector<int>& inorder) {
unordered_map<int, size_t> in_entry_idx_map;
for (size_t i = 0; i < inorder.size(); ++i) {
in_entry_idx_map.emplace(inorder[i], i);
}
return ReconstructPreInOrdersHelper(preorder, 0, preorder.size(), inorder, 0, inorder.size(),
in_entry_idx_map);
}
// Reconstructs the binary tree from pre[pre_s : pre_e - 1] and
// in[in_s : in_e - 1].
TreeNode *ReconstructPreInOrdersHelper(const vector<int>& preorder, size_t pre_s, size_t pre_e,
const vector<int>& inorder, size_t in_s, size_t in_e,
const unordered_map<int, size_t>& in_entry_idx_map) {
if (pre_s == pre_e || in_s == in_e) {
return nullptr;
}
auto idx = in_entry_idx_map.at(preorder[pre_s]);
auto left_tree_size = idx - in_s;
auto node = new TreeNode(preorder[pre_s]);
node->left = ReconstructPreInOrdersHelper(preorder, pre_s + 1, pre_s + 1 + left_tree_size,
inorder, in_s, idx, in_entry_idx_map);
node->right = ReconstructPreInOrdersHelper(preorder, pre_s + 1 + left_tree_size, pre_e,
inorder, idx + 1, in_e, in_entry_idx_map);
return node;
}
};
类似题目：
[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树
