Binary Tree Postorder Traversal--LeetCode

1.题目

Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?

2.题意

二叉树后序遍历

3.分析

1）递归，直接按照左、右、root的顺序放进vector即可

2）借助两个栈
s1将每两个孩子节点按照左、右的顺序进栈
则每个栈顶元素出栈时再按照root、右、左的顺序进入s2
将s2所有元素出栈到vector即可得到左、右、root的后序遍历序列

4.代码

1）递归版

class Solution {public:    vector<int> postorderTraversal(TreeNode* root) {        vector<int> result;        postorder(result, root);        return result;    }private:    void postorder(vector<int> &result, TreeNode *root) {        if(root == nullptr)            return;        postorder(result, root->left);        postorder(result, root->right);        result.push_back(root->val);    }};

2）迭代版

class Solution {public:    vector<int> postorderTraversal(TreeNode* root) {        vector<int> result;        if(root == nullptr)            return result;        stack<const TreeNode*> s1;        stack<const TreeNode*> s2;        const TreeNode *p = root;        s1.push(root);        while(!s1.empty())        {            p = s1.top();            s1.pop();            s2.push(p);            if(p->left)                s1.push(p->left);            if(p->right)                s1.push(p->right);        }        while(!s2.empty())        {            p = s2.top();            s2.pop();            result.push_back(p->val);        }        return result;    }};