LeetCode 105. 从前序与中序遍历序列构造二叉树

105. 从前序与中序遍历序列构造二叉树

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

递归解法

class Solution {
public:
    TreeNode* buildTree(vector& preorder, vector& inorder) {
        for (int i = 0; i < inorder.size(); ++i) {
            mp[inorder[i]] = i;
        }
        return myBuildTree(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
    }
private:
    map mp;
    TreeNode* myBuildTree(vector& preorder, vector& inorder, int preLeft, int preRight, int inLeft, int inRight) {
        if (preLeft > preRight) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(preorder[preLeft]);
        int inRoot = mp[preorder[preLeft]];
        int subLeft = inRoot - inLeft;
        root->left = myBuildTree(preorder, inorder, preLeft + 1, preLeft + subLeft, inLeft, inRoot - 1);
        root->right = myBuildTree(preorder, inorder, preLeft + subLeft + 1, preRight, inRoot + 1, inRight);
        return root;
    }
};

迭代方法，太费脑力

class Solution {
public:
    TreeNode* buildTree(vector& preorder, vector& inorder) {
        stack stk;
        TreeNode *root = new TreeNode(preorder[0]);
        stk.push(root);
        int inorderIndex = 0;
        for (int i = 1; i < preorder.size(); ++i) {
            int preoderVal = preorder[i];
            TreeNode* node = stk.top();
            if (stk.top()->val != inorder[inorderIndex]) {
                node->left = new TreeNode(preoderVal);
                stk.push(node->left);
            } else {
                while (stk.size() && stk.top()->val == inorder[inorderIndex]) {
                    node = stk.top();
                    stk.pop();
                    ++inorderIndex;
                }
                node->right = new TreeNode(preoderVal);
                stk.push(node->right);
            }
        }
        return root;
    }
};