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

二叉树的后续遍历

 

1、递归版本

/**
 * 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。。。为什么我总是CE呢。。因为写完之后觉得程序太简单，所以不想检查。。

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;
    }
};

后续遍历需要两个指针来记录状态now和pre，还有3个循环的边界，感觉这个程序可以写成很多不同的版本。

关键是如何判断now这个点的左子树和右子树都已经被访问过了。