Binary Tree Postorder Traversal

标签：leetcode

问题描述

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?

解决方案

/**
 * 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) {
        if( root == nullptr ) {
            return { };
        }    
        vector<int> result;
        stack<TreeNode*> nodeStack;
        nodeStack.push( root );
        
        while( !nodeStack.empty() ) {
            auto t = nodeStack.top();
            nodeStack.pop();
            if( t != nullptr ) {
                result.push_back( t->val );
                nodeStack.push( t->right );
                nodeStack.push( t->left );
            }
        }
        reverse( result.begin(), result.end() );
        return result;
    }
};

Binary Tree Postorder Traversal

标签：leetcode

原文地址：http://blog.csdn.net/senlinzm/article/details/39011767