[Leetcode][Tree][Binary Tree Postorder Traversal]

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二叉樹的後續遍曆

 

1、遞迴版本

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void dfsPostorderTraversal(TreeNode *now, vector<int> &result) {        if (now == NULL) {            return;        }        dfsPostorderTraversal(now->left, result);        dfsPostorderTraversal(now->right, result);        result.push_back(now->val);    }    vector<int> postorderTraversal(TreeNode *root) {        vector<int> result;        dfsPostorderTraversal(root, result);        return result;    }};

　　CE一次。。。為什麼我總是CE呢。。因為寫完之後覺得程式太簡單，所以不想檢查。。

2、迭代版本

/** * Definition for binary tree * 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> result;        TreeNode *now, *pre;        stack<TreeNode*> s;        now = root;        pre = NULL;        //while (now != NULL || !s.empty()) {        do {            while (now != NULL) {                s.push(now);                now = now->left;            }            pre = NULL;            while (!s.empty()) {                now = s.top();                if (now->right != pre) {                    now = now->right;                    break;                } else {                    result.push_back(now->val);                    pre = now;                    s.pop();                }            }        } while(!s.empty());        return result;    }};

後續遍曆需要兩個指標來選項組now和pre，還有3個迴圈的邊界，感覺這個程式可以寫成很多不同的版本。

關鍵是如何判斷now這個點的左子樹和右子樹都已經被訪問過了。