題目描述

給定一個二叉樹，返回它的 後序 遍歷。

示例:

輸入: [1,null,2,3]  
   1
    \
     2
    /
   3 
輸出: [3,2,1]

進階: 遞迴演算法很簡單，你可以通過迭代演算法完成嗎？

解題思路1：
用兩個棧和一個數組。

/**
 * 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) {
        stack<TreeNode*> s1;
        stack<TreeNode*> s2;
        vector<int> res;
        if(!root){
            return res;
        }
        s1.push(root);
        TreeNode* p=root;
        while
 
(!s1.empty()){
            p=s1.top();
            s1.pop();
            s2.push(p);
            if(p->left!=NULL){
                s1.push(p->left);
            }
            if(p->right!=NULL){
                s1.push(p->right);
            }
        }
        while(!s2.empty()){
            p=
 
s2.top();
            res.push_back(p->val);
            s2.pop();
        }
        return res;
    }
};

解題思路2：
用一個棧和一個數組。

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

遞迴：

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