LeetCode刷题笔记(树):binary-tree-postorder-traversal



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题目描述

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

For example:
Given binary tree{1,#,2,3},

1
\
2
/
3

return[3,2,1].

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

给定一个二叉树,返回其节点值的后序遍历。

注意:递归解决方案是微不足道的,你可以迭代地做?

解题思路

两种方法,递归和迭代,递归太简单了,直接贴出代码就不多说了。

后序遍历的非递归版可以说是三种遍历中最难的了,因为要先穷举左右结点,最后才是根结点。在外循环中,第一步还是依次将左子树入栈;然后看栈顶元素,如果栈顶元素没有右孩子,或者右孩子时候已经遍历过,那么就弹出栈顶元素,并记录;否则,遍历右子树。

C++版代码实现

递归

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/**
* 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 postorder(TreeNode *root, vector<int> &res){
if(root){
postorder(root->left, res);
postorder(root->right, res);
res.push_back(root->val);
}
}

vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
postorder(root, res);
return res;
}
};

迭代

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

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完的汪(∪。∪)。。。zzz

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