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

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / 
  9  20
    /  
   15   7

给一棵树的先序和中序遍历，构建二叉树。

Java:

public TreeNode buildTree(int[] preorder, int[] inorder) {
    return helper(0, 0, inorder.length - 1, preorder, inorder);
}

public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
    if (preStart > preorder.length - 1 || inStart > inEnd) {
        return null;
    }
    TreeNode root = new TreeNode(preorder[preStart]);
    int inIndex = 0; // Index of current root in inorder
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == root.val) {
            inIndex = i;
        }
    }
    root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
    root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
    return root;
}

Python: Recursive

def buildTree(self, preorder, inorder):
    if inorder:
        ind = inorder.index(preorder.pop(0))
        root = TreeNode(inorder[ind])
        root.left = self.buildTree(preorder, inorder[0:ind])
        root.right = self.buildTree(preorder, inorder[ind+1:])
        return root

Python:

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    # @param preorder, a list of integers
    # @param inorder, a list of integers
    # @return a tree node
    def buildTree(self, preorder, inorder):
        lookup = {}
        for i, num in enumerate(inorder):
            lookup[num] = i
        return self.buildTreeRecu(lookup, preorder, inorder, 0, 0, len(inorder))

    def buildTreeRecu(self, lookup, preorder, inorder, pre_start, in_start, in_end):
        if in_start == in_end:
            return None
        node = TreeNode(preorder[pre_start])
        i = lookup[preorder[pre_start]]
        node.left = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1, in_start, i)
        node.right = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1 + i - in_start, i + 1, in_end)
        return node

if __name__ ==  "__main__":
    preorder = [1, 2, 3]
    inorder = [2, 1, 3]
    result = Solution().buildTree(preorder, inorder)
    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] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树

　

All LeetCode Questions List 题目汇总