[LeetCode] Binary Tree Postorder题解

Binary Tree Postorder

Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

这是一道LeetCode中标记为Hard的题。事实上如果没有限定不使用递归的话，这道题是非常简单的。所以我只简单回顾一下这道题的两种解法：递归和迭代。

递归法实现后序遍历

算法复杂度为O（n）。

class Solution {
public:vector<int> postorderTraversal(TreeNode* root) {vector<int> re;print(root,re);return re;}void print(TreeNode *node,vector<int> &re){if(node == NULL) return;       print(node->left,re);//左    print(node->right,re);//右re.push_back(node->val);//中}
};

递归实现前序遍历和后序遍历，只要把print函数中“左右中”三行代码改成相应的顺序即可。

迭代实现后序遍历

迭代实现遍历的本质是广度优先搜索，思路如下：

Create an empty stack, Push root node to the stack.

Do following while stack is not empty.

pop an item from the stack and print it.

push the left child of popped item to stack.

push the right child of popped item to stack.

reverse the ouput.

其中，容易搞错的是输出“中”后，要先push左节点，再push右节点。因为对栈来说，先进去的左节点会后输出（先进后出，后进先出），就实现了“中右左”的顺序，再反转（reverse）就得到了后续遍历（左右中）。

算法复杂度为O（n）。

class Solution {
public:vector<int> postorderTraversal(TreeNode* root) {vector<int> re;stack<TreeNode*> visit;if(root != NULL) visit.push(root);while(!visit.empty()){TreeNode *topNode = visit.top();visit.pop();//top方法只是获取最上面的元素，所以要用pop方法弹出re.push_back(topNode->val);if(topNode->left != NULL)visit.push(topNode->left);if(topNode->right != NULL)visit.push(topNode->right);}reverse(re.begin(),re.end());return re;}
};

转载于:https://www.cnblogs.com/liangf27/p/9356871.html

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[LeetCode] Binary Tree Postorder题解

递归法实现后序遍历

迭代实现后序遍历

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