Leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/ruihuank/article/details/78998489

本文介绍了一种使用前序遍历和中序遍历构建二叉树的方法。通过递归方式和栈实现两种算法来重建原始二叉树结构，并详细解释了每一步骤。

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

/**
 * 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:
    using CIt = vector<int>::const_iterator;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return rebuild(preorder.begin(), preorder.end(), inorder.begin(), inorder.end());
    }

private:

    // b1 and e1 are for preorder
    TreeNode* rebuild(CIt b1, CIt e1, CIt b2, CIt e2) {
        if (e1 == b1) return nullptr;
        int val = *b1;
        TreeNode* ret = new TreeNode(val);
        CIt it = find(b2, e2, val);

        ret->left = rebuild(b1 + 1, b1 + 1 + (it - b2), b2, it);
        ret->right  = rebuild(b1 + 1 + (it - b2),e1, it + 1, e2);
        return ret;
    }
};

/**
 * 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) {
        if (preorder.size() == 0) return nullptr;
        stack<TreeNode*> s;
        unordered_map<int, int> m;
        for (size_t i = 0; i != inorder.size(); ++i) {
            m[inorder[i]] = i;
        }
        TreeNode* root = new TreeNode(preorder[0]);
        s.push(root);
        TreeNode* p = root;

        size_t id = 1;

        while (id != preorder.size()) {

            if (m[preorder[id]] < m[p->val]) {
                p->left = new TreeNode(preorder[id]);
                s.push(p->left);
                p = p->left;
                ++id;
            } else {
                // 当前 root
                s.pop();

                // bijiao上一个 root
                if (s.empty() || m[preorder[id]] < m[s.top()->val]){
                    p->right = new TreeNode(preorder[id]);
                    s.push(p->right);
                    p = p->right;
                    ++id;
                } else {
                    p = s.top();
                }  
            }       
        }
        return root;


    }
};