LeetCode 题解（86）: Binary Tree Postorder Traversal

题目：

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?

题解：

考察数据结构，一个set加一个堆栈解决问题。

c++：

/** * Definition for a binary tree node. * 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) {        unordered_set<TreeNode*> visited;        stack<TreeNode*> nodeStack;                vector<int> result;        if(root == NULL)            return result;        nodeStack.push(root);        while(nodeStack.size() != 0) {            TreeNode* current = nodeStack.top();            if(current->right != NULL && visited.find(current->right) == visited.end())                nodeStack.push(current->right);            if(current->left != NULL && visited.find(current->left) == visited.end())                nodeStack.push(current->left);            if(current->right == NULL && current->left == NULL) {                result.push_back(current->val);                nodeStack.pop();                visited.insert(current);            }            if((current->right != NULL && visited.find(current->right) != visited.end()) || (current->left != NULL && visited.find(current->left) != visited.end())) {                result.push_back(current->val);                nodeStack.pop();                visited.insert(current);            }        }        return result;    }    };

Java版：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        public List<Integer> postorderTraversal(TreeNode root) {        Set<TreeNode> visited = new HashSet<TreeNode>();        List<Integer> result = new ArrayList<>();        if(root == null)            return result;        Stack<TreeNode> nodeStack = new Stack<TreeNode>();        nodeStack.push(root);        while(nodeStack.empty() == false) {            TreeNode current = nodeStack.peek();            if(current.right != null && visited.contains(current.right) == false) {                nodeStack.push(current.right);            }            if(current.left != null && visited.contains(current.left) == false) {                nodeStack.push(current.left);            }            if(current.left == null && current.right == null) {                nodeStack.pop();                visited.add(current);                result.add(current.val);            }            if((current.right != null && visited.contains(current.right) == true) || (current.left != null && visited.contains(current.left) == true)){                nodeStack.pop();                visited.add(current);                result.add(current.val);            }        }        return result;    }}

Python版：

class Solution:    # @param {TreeNode} root    # @return {integer[]}    def postorderTraversal(self, root):        result = []        if root == None:            return result        visited = []        nodeStack = []        nodeStack.append(root)        while len(nodeStack) != 0:            current = nodeStack[-1]            if current.right != None and not (current.right in visited):                nodeStack.append(current.right)            if current.left != None and not (current.left in visited):                nodeStack.append(current.left)            if current.left == None and current.right == None:                nodeStack.pop()                result.append(current.val)                visited.append(current)            if (current.right != None and (current.right in visited)) or (current.left != None and (current.left in visited)):                nodeStack.pop()                result.append(current.val)                visited.append(current)        return result