标签：blog os for leetcode io res
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。
[Leetcode][Tree][Binary Tree Postorder Traversal],布布扣,bubuko.com
[Leetcode][Tree][Binary Tree Postorder Traversal]
标签：blog os for leetcode io res
原文地址：http://www.cnblogs.com/poemqiong/p/3813730.html
