[C++进阶]二叉树进阶的一些面试题(二)

144. 二叉树的前序遍历

这题很简单,递归代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void preorder(TreeNode *root, vector &res) 
    {
        if (root == nullptr) 
        {
            return;
        }
        res.push_back(root->val);
        preorder(root->left, res);
        preorder(root->right, res);
    }

    vector preorderTraversal(TreeNode *root) 
    {
        vector res;
        preorder(root, res);
        return res;
    }
};

但是这道题我们的目的是用迭代做出来

思路

1.先访问左路结点
2.再访问左路结点的右子树
这里访问右子树要以循环从栈依次取出这些结点,循环子问题的思想访问左路结点的右子树。
(中序和后序跟前序的思路完全一 致，只是前序先访问根还是后访问根的问题。后序稍微麻烦- -些，因为后序遍历的顺序是左子树右子树根，当取到左路结点的右子树时，需要想办法标记判断右子树是否访问过了，如果访问过了,就直接访问根，如果访问过需要先访问右子树

代码示例:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector preorderTraversal(TreeNode* root) 
    {
        vector res;
        if (root == nullptr) 
        {
            return res;
        }
        stack stk;
        TreeNode* cur = root;
        while (!stk.empty() || cur) 
        {
        //每次循环开始代码访问一棵树的开始
        //访问左路节点,左路节点入栈

            while (cur) 
            {
                res.push_back(cur->val);
                stk.push(cur);
                cur = cur->left;
            }
            //取一个左路节点的右子树出来访问
            cur = stk.top();
            stk.pop();

            cur = cur->right;
        }
        return res;
    }
};

94. 二叉树的中序遍历

这题是上面那题的姊妹题,换换顺序就能写好了

迭代写法:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector inorderTraversal(TreeNode* root) 
    {
        vector res;
        if (root == nullptr) 
        {
            return res;
        }
        stack stk;
        TreeNode* cur = root;
        while (!stk.empty() || cur) 
        {
            while (cur) 
            {
                stk.push(cur);
                cur = cur->left;
            }
            cur = stk.top();
            stk.pop();
            res.push_back(cur->val);
            cur = cur->right;
        }
        return res;
    }
};

145. 二叉树的后序遍历

一样姊妹题:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector postorderTraversal(TreeNode* root) {
       vector res;
        if (root == nullptr) 
        {
            return res;
        }
        stack stk;
        TreeNode* cur = root;
        TreeNode* prev = nullptr;
        while (!stk.empty() || cur) 
        {
            while (cur) 
            {
                stk.push(cur);
                cur = cur->left;
            }
            cur = stk.top();
            stk.pop();
            if(cur->right==nullptr || cur->right == prev)
            {
                res.push_back(cur->val);
                prev = cur;
                cur = nullptr;
            }
            else
            {
                stk.push(cur);
                cur=cur->right;
            }
        }
        return res;
    }
};