二叉树与双向链表问题

做完这道题，我深刻认识到通过边画图边写代码揣摩分析才能真正弄懂这道题，而且如果不是独立思考AC看了题解恐怕就很难弄懂这道题。

这里整理一下题解：本题思路有两种：递归和非递归

Java版AC代码：

数据结构定义：

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }
}
*/

递归版：

public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null)
            return null;
        TreeNode res = pRootOfTree;
        if(pRootOfTree.right != null) {
            TreeNode tRight = Convert(pRootOfTree.right);
            pRootOfTree.right = tRight;
            tRight.left = pRootOfTree;
        }
             
        if(pRootOfTree.left != null) {
            TreeNode tLeft = Convert(pRootOfTree.left);
            TreeNode tmp = tLeft;
            while(tmp.right != null) {
                tmp = tmp.right;
            }
            tmp.right = pRootOfTree;
            pRootOfTree.left = tmp;
            res = tLeft;
        }
        return res;
}

非递归版：

Java中有关Stack的API

import java.util.Stack;
public class Solution {//画图容易理解
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null)
            return null;
        Stack<TreeNode> st = new Stack<TreeNode>();
        TreeNode tnp, tmp = pRootOfTree;
        TreeNode res = tmp;
        st.push(tmp);
        while(tmp.left != null) {
            tmp = tmp.left;
            st.push(tmp);
        }
        
        if(!st.empty()) 
            res = st.peek();
        
        while(!st.empty()) {
            tmp = st.pop();
            tnp = tmp.right;
            if(tnp != null) {
                st.push(tnp);
                while(tnp.left != null) {
                    tnp = tnp.left;
                    st.push(tnp);
                }
                tnp.left = tmp;
                tmp.right = tnp;
            } else {
                if(!st.empty()) {//peek()操作前一定要进行判空操作，否则会产生EmptyStack异常
                    tmp.right = st.peek();
                    tmp.right.left = tmp;
                }
            }
        }
        return res;
    }
}

C++版AC代码：

数据结构：

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/

递归版：

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL)
            return NULL;
        TreeNode* res = pRootOfTree;
        TreeNode* tmp = res;
        if(tmp->right != NULL) {
            TreeNode* tRight = Convert(tmp->right);
            pRootOfTree->right = tRight;
            tRight->left = pRootOfTree;
        }
        
        if(tmp->left != NULL) {
            TreeNode* tLeft = Convert(tmp->left);
            res = tLeft;
            while(tLeft->right != NULL) {
                tLeft = tLeft->right;
            }
            tLeft->right = pRootOfTree;
            pRootOfTree->left = tLeft;
        }
        
        return res;
    }
};

非递归版：

C++关于stack(首字母小写）的API

1. bool empty();//判空
2. void pop();//出栈
3. void push(const Type &val);//入栈
4. size_type size();//得到栈的大小
5. Type &top();//取得栈顶元素引用但不弹出

#include <bits/stdc++.h>
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL)
            return NULL;
        stack<TreeNode*> st;
        st.push(pRootOfTree);
        TreeNode* tmp = pRootOfTree;
        TreeNode* tnp, *res = tmp;
        while(tmp->left != NULL) {
            tmp = tmp->left;
            st.push(tmp);
        }
        
        if(!st.empty()) {
            res = st.top();
        }
        
        while(!st.empty()) {
            tmp = st.top();
            st.pop();
            tnp = tmp->right;
            if(tnp != NULL) {
                st.push(tnp);
                while(tnp->left != NULL) {
                    tnp = tnp->left;
                    st.push(tnp);
                }
                tnp->left = tmp;
                tmp->right = tnp;
            } else {
                if(!st.empty()) {//一定注意要判空
                    tmp->right = st.top();
                    tmp->right->left = tmp;
                }
            }
        }
        return res;
    }
};