leetcode：二叉树之Binary Tree Postorder Traversal

leetcode：二叉树之Binary Tree Postorder Traversal

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?

即，二叉树后序遍历。

c++实现：

#include <iostream>#include <vector>#include <malloc.h>#include <stack>using namespace std;struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) { }};void CreateBiTree(TreeNode* &T){char ch;cin>>ch;if(ch=='#') T=NULL;else{T = (TreeNode*)malloc(sizeof(TreeNode));if(!T) exit(0);T->val = ch-'0';CreateBiTree(T->left);  CreateBiTree(T->right); }}vector<int> postorderTraversal(TreeNode *root){vector<int> result;/* p，正在访问的结点，q，刚刚访问过的结点*/const TreeNode *p, *q;stack<const TreeNode *> s;p = root;do {while (p != NULL) /* 往左下走*/{ s.push(p);p = p->left;}q=NULL;while (!s.empty()) {p = s.top();s.pop();/* 右孩子不存在或已被访问，访问之*/if (p->right == q) {result.push_back(p->val);q = p; /* 保存刚访问过的结点*/} else {/* 当前结点不能访问，需第二次进栈*/s.push(p);/* 先处理右子树*/p = p->right;break;}}} while (!s.empty());return result;}int main(){TreeNode* root(0);CreateBiTree(root);cout<<"输出后序遍历结果："<<endl;vector<int> v;v=postorderTraversal(root);for(int i = 0;i < v.size();i++)        cout<<v[i];cout<<endl;    return 0;}

测试结果：