【LeetCode】一文详解二叉树的三大遍历：前序、中序和后序（python和C++实现）

深度学习技术前沿公众号博主

发布于 2020-05-18 10:52:52

7790

发布于 2020-05-18 10:52:52

文章被收录于专栏：深度学习技术前沿深度学习技术前沿深度学习技术前沿

本文主要包括利用递归和栈的方法实现二叉树的前序、中序、后序遍历！

144. 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]

输出: [1,2,3]

解题思路

1.1 树的前序遍历--非递归方法（栈）

因为先访问根节点，所以直接将root的val放入答案（ans）容器
利用stack来储存root。
当左子树遍历完后，取出root接着遍历右子树。

C++实现：

/**
 * 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) {
        std::stack<TreeNode*> stack;
        vector<int> ans;
        while(root != NULL || !stack.empty()) {
            while(root != NULL) {
                stack.push(root);
                ans.push_back(root->val);
                root = root->left;
            }
            root = stack.top();
            stack.pop();
            root = root->right;
        }
        return ans;
    }
};

Python实现:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        if root is None:
            return []
        stack,res=[root,],[]
        while stack:
            root = stack.pop()
            if root is not None:
                res.append(root.val)
                if root.right is not None:
                    stack.append(root.right)
                if root.left is not None:
                    stack.append(root.left)
        return res

1.2 递归的思想

C++实现

class Solution {
public:
    vector<int> ans;
    vector<int> preorderTraversal(TreeNode* root) {
        if(root != NULL){
            ans.push_back(root -> val);
            preorderTraversal(root -> left);
            preorderTraversal(root -> right);
        }
        return ans;
    }
};

Python实现：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        def preorder(root):
            if not root:return 
            res.append(root.val)
            preorder(root.left)
            preorder(root.right)
        preorder(root)
        return res

94. 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]

输出: [1,2,3]

解题思路

2.1 利用栈的思想

C++实现：

/**
 * 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) {
        vector<int> res;
        stack<TreeNode*> stack;
        //s.push(root);
        while(root || s.size() > 0){
            while(root){
                stack.push(root);
                root = root->left;
            }
            root = stack.top(); 
            stack.pop();
            res.push_back(root->val);
            root = root->right;
        }
        return res;
    }
};

python实现：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        stack = []
        while root or stack:
            # 把左子树压入栈中
            while root:
                stack.append(root)
                root = root.left
            # 输出 栈顶元素
            root = stack.pop()
            res.append(root.val)
            # 遍历右子树
            root = root.right
        return res

2.2 递归的思想

C++实现：

/**
 * 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> ans;
    vector<int> inorderTraversal(TreeNode* root) {
        if(root != NULL){
            inorderTraversal(root -> left);
            ans.push_back(root -> val);
            inorderTraversal(root -> right);
        }
        return ans;
    }
};

Python实现：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        def inorder(root):
            if not root:return 
            inorder(root.left)
            res.append(root.val)
            inorder(root.right)
        inorder(root)
        return res

145. 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]

输出: [1,2,3]

解题思路

3.1 利用迭代的思想（栈）

C++实现：

先遍历左节点直到左节点为null。
开始遍历右节点，若该右节点有左节点，优先遍历左节点。
使用rightchild来记录右节点是否已被遍历过。若是：则说明以该点为根的子树已被遍历，输出根节点。若否：就开始遍历右节点，回到第二步。

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

Python实现：

从根节点开始依次迭代，弹出栈顶元素输出到输出列表中，然后依次压入它的所有孩子节点，按照从上到下、从左至右的顺序依次压入栈中。
因为深度优先搜索后序遍历的顺序是从下到上、从左至右，所以需要将输出列表逆序输出。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        if root is None:
            return []

        stack, output = [root, ], []
        while stack:
            root = stack.pop()
            output.append(root.val)
            if root.left is not None:
                stack.append(root.left)
            if root.right is not None:
                stack.append(root.right)
                
        return output[::-1]

3.2 递归的思想

C++实现：

/**
 * 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> ans;
    vector<int> postorderTraversal(TreeNode* root) {
        if(root != NULL){
            postorderTraversal(root->left);
            postorderTraversal(root->right);
            ans.push_back(root -> val);
        }
        return ans;
    }
};

Python实现：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        def postorder(root):
            if not root:
                return 
            postorder(root.left)
            postorder(root.right)
            res.append(root.val)
        postorder(root)
        return res

本文参与腾讯云自媒体分享计划，分享自微信公众号。

原始发表：2020-04-16，如有侵权请联系 cloudcommunity@tencent.com 删除

二叉树