二叉树的后序遍历（非递归）

题目描述

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?

Solution 1（非递归实现）

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    //前序遍历是根->左->右放入栈中出栈顺序根->右->左，然后逆序reverse
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int>res;
        if(root==nullptr) return res;
        stack<TreeNode*>st;
        st.push(root);
        while(!st.empty())
        {
            TreeNode *temp=st.top();
            st.pop();
            res.push_back(temp->val);
            if(temp->left!=nullptr)
                st.push(temp->left);
            if(temp->right!=nullptr)
                st.push(temp->right);
        }
        reverse(res.begin(),res.end());
        return res;
    }
};

Solution 2（递归实现）

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

同样可以应用到前序遍历上