[每日leetcode]二叉树的前序、中序、后序遍历

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二叉树的前序遍历

给定一个二叉树，返回它的 前序 遍历。
示例:

输入: [1,null,2,3]  
  1
   \
    2
   /
  3 

输出: [1,2,3]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

c++代码

递归

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> vals;
        preorder(root,vals);
        return vals;
    }
    void preorder(TreeNode* root,vector<int>& vals)
    {
        if(root==NULL)
            return ;
        vals.push_back(root->val);
        preorder(root->left,vals);
        preorder(root->right,vals);
    }
};

迭代

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        if(root==NULL)
            return {};
        vector<int> vals;
        stack<TreeNode*> s;
        TreeNode *cur=root;
        while(cur||!s.empty())
        {
            if(cur)
            {
                s.push(cur);//根节点先入栈，
                vals.push_back(cur->val);
                //左子树入栈
                cur=cur->left;
            }
            else
            {
                cur=s.top()->right;
                s.pop();
            }
            
        }
        return vals;
    }
};

——————————————

二叉树的中序遍历

给定一个二叉树，返回它的中序 遍历。
示例:

输入: [1,null,2,3]
   1
    \
     2
    /
   3
输出: [1,3,2]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

c++代码

递归

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> vals;
        preorder(root,vals);
        return vals;
    }
    void preorder(TreeNode* root,vector<int>& vals)
    {
        if(root==NULL)
            return ;
        
        preorder(root->left,vals);
        vals.push_back(root->val);
        preorder(root->right,vals);
    }
};

迭代

/**
 * 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:
    vector<int> inorderTraversal(TreeNode* root) {
        if(root==NULL)
            return {};
        vector<int> vals;
        stack<TreeNode*> s;
        TreeNode *cur=root;
        while(cur||!s.empty())
        {
            if(cur)
            {
                s.push(cur);
                cur=cur->left;
            }
            else
            {
                cur=s.top()->right;
                vals.push_back(s.top()->val);
                s.pop();
            }
        }
        return vals;
    }
};

——————————————

二叉树的后序遍历

给定一个二叉树，返回它的 后序 遍历。
示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

c++代码

递归

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> vals;
        preorder(root,vals);
        return vals;
    }
    void preorder(TreeNode* root,vector<int>& vals)
    {
        if(root==NULL)
            return ;
        
        preorder(root->left,vals);       
        preorder(root->right,vals);
        vals.push_back(root->val);
    }
};

迭代

思路：正常顺序是先走左子树，再走右子树，再走根。直接先走根，再走右子树，再走左子树最后将结果reverse比较简单。

/**
 * 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:
    vector<int> postorderTraversal(TreeNode* root) {
        if(root==NULL)
            return {};
        vector<int> vals;

        stack<TreeNode*> s;
        TreeNode* cur = root;

        while (cur || !s.empty()) {
            if (cur) {
                s.push(cur);
                vals.push_back(cur->val);
                cur = cur->right;
            } else {
                cur = s.top()->left;
                s.pop();
            }
        }
        reverse(vals.begin(),vals.end());
        return vals;
    }
};