889.Construct Binary Tree from Preorder and Postorder Traversal

Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

Example 1:

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Note:

• 1 <= pre.length == post.length <= 30
• pre[] and post[] are both permutations of 1, 2, ..., pre.length.
• It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

Runtime: 8 ms, faster than 98.12% of C++ online submissions for Construct Binary Tree from Preorder and Postorder Traversal.

#include<stdlib.h>
#include<vector>
#include<stack>
#include<queue>
#include <iostream>

using namespace std;

//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 *constructFromPrePost(vector<int> &pre, vector<int> &post) {
        if (pre.size() == 0) return nullptr;
        stack<TreeNode *> st;
        TreeNode *root = new TreeNode(pre[0]);
        st.push(root);
        int j = 0;
        int i = 0;
        TreeNode *node = nullptr;
        for(int i=1;i<pre.size();++i){
            while (st.top()->val == post[j]) {
                node = st.top();
                st.pop();
                printf("pop %d\n",node->val);
                if (st.top()->left == nullptr) {
                    st.top()->left = node;
                    printf("%d left child is %d\n", st.top()->val, node->val);
                } else {
                    st.top()->right = node;
                    printf("%d right child is %d\n", st.top()->val, node->val);
                }
                j++;
                printf("j: %d\n",j);
            }
            if (i < pre.size()){
                st.push(new TreeNode(pre[i]));
                printf("push %d\n",pre[i]);
            }
        }

        while (true) {
            node = st.top();
            st.pop();
            if(st.empty()) break;
            printf("pop %d\n",node->val);
            if (st.top()->left == nullptr) {
                st.top()->left = node;
                printf("%d left child is %d\n", st.top()->val, node->val);
            } else {
                st.top()->right = node;
                printf("%d right child is %d\n", st.top()->val, node->val);
            }
            j++;
            printf("j: %d\n",j);
        }
        return root;
//        printf("return\n");
//        printf("root->value %d\n",root->val);
//        return root;
    }
};

void show_tree(TreeNode *root) {
    if (root == nullptr) {
        cout<<"root is nullptr"<<endl;
        return;
    }
    cout<<"root is not nullptr"<<endl;
    queue<TreeNode *> qu;
    qu.push(root);
    int sz;
    while (!qu.empty()) {
        sz = qu.size();
        TreeNode* node= nullptr;
        for (int i = 0; i < sz; ++i) {
            node=qu.front();
            cout << node->val << " ";
            qu.pop();
            if (node->left)
                qu.push(node->left);
            if (node->right)
                qu.push(node->right);
        }
        cout << "\n";
    }
}

int main() {
    vector<int> pre{1, 2, 4, 5, 3, 6, 7};
    vector<int> post{4, 5, 2, 6, 7, 3, 1};
    Solution solution;
    TreeNode *res = solution.constructFromPrePost(pre, post);

//    solution.constructFromPrePost(pre, post);
    printf("res value %d\n",res->val);
    show_tree(res);
    return 0;

}

提交代码

class Solution {
public:
    TreeNode *constructFromPrePost(vector<int> &pre, vector<int> &post) {
        if (pre.size() == 0) return nullptr;
        stack<TreeNode *> st;
        TreeNode *root = new TreeNode(pre[0]);
        st.push(root);
        int j = 0;
        TreeNode *node = nullptr;
        for(int i=1;i<=pre.size();++i){
            while (st.top()->val == post[j]) {
                node = st.top();
                st.pop();
                //printf("pop %d\n",node->val);
                if(st.empty())
                    return root;
                if (st.top()->left == nullptr) {
                    st.top()->left = node;
                    //printf("%d left child is %d\n", st.top()->val, node->val);
                } else {
                    st.top()->right = node;
                    //printf("%d right child is %d\n", st.top()->val, node->val);
                }
                j++;
                //printf("j: %d\n",j);
            }
            if (i < pre.size()){
                st.push(new TreeNode(pre[i]));
                //printf("push %d\n",pre[i]);
            }
        }
    }
};

原文地址：https://www.cnblogs.com/learning-c/p/9847610.html