【小白刷剑指offer】第四题 重建二叉树 c++/python

4 重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

❤️递归方法和一些题目分析

运用栈实现

# python 解法
class Solution:
    def buildTree(self, preorder, inorder):
        if not preorder:
            return None
        root = TreeNode(preorder[0])
        stack = [root]
        inorderIndex = 0
        for i in range(1, len(preorder)):
            addnode = TreeNode(preorder[i])
            head = stack[-1]
            if stack[-1].val != inorder[inorderIndex]:
                head.left = addnode
                stack.append(addnode)
            else:
                while stack and stack[-1].val == inorder[inorderIndex]:
                    rootnode = stack.pop()
                    inorderIndex += 1  
                rootnode.right = addnode
                stack.append(addnode)
        return root

/**
 * 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 or inorder.size() == 0 or preorder.size() != inorder.size()){
    
    
            return 0;
        }
        stack<TreeNode*> st;
        TreeNode* root = new TreeNode(preorder[0]);
        st.push(root);
        int j = 0;
        for(int i=1;i<preorder.size();i++){
    
    
            TreeNode* now = new TreeNode(preorder[i]);
            TreeNode* head = st.top();
            if(head->val == inorder[j]){
    
    
                while (not st.empty() && j<preorder.size() && st.top()->val == inorder[j]){
    
    
                    head = st.top();
                    st.pop();
                    j = j+1;
                    }
                head->right = now;
                }
            else{
    
    
                head->left = now;
            }
            st.push(now);
        }
    return root;
    }
};