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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].
C++ 实现代码：
#include<iostream>#include<new>#include<vector>#include<stack>using namespace std;//Definition for binary treestruct TreeNode{ int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution{public: vector<int> postorderTraversal(TreeNode *root) { stack<TreeNode*> st; vector<int> ret; if(root==NULL) return vector<int>(); st.push(root); TreeNode *pre=NULL; TreeNode *cur; while(!st.empty()) { cur=st.top(); if((cur->left==NULL&&cur->right==NULL)||((pre!=NULL)&&(pre==cur->left||pre==cur->right))) { ret.push_back(cur->val); st.pop(); pre=cur; } else { if(cur->right) st.push(cur->right); if(cur->left) st.push(cur->left); } } return ret; } void createTree(TreeNode *&root) { int i; cin>>i; if(i!=0) { root=new TreeNode(i); if(root==NULL) return; createTree(root->left); createTree(root->right); } }};int main(){ Solution s; TreeNode *root; s.createTree(root); vector<int> vec; vec=s.postorderTraversal(root); for(auto a:vec) cout<<a<<" "; cout<<endl;}
Binary Tree Postorder Traversal
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