[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;
    }
};

　

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[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树

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