标签:modal tor col 一个 || iss des else color
给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-postorder-traversal
迭代:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> postorderTraversal(TreeNode* root) { 13 vector<int> ans; 14 if(!root) return ans; 15 stack<TreeNode*> nodeStack1,nodeStack2; 16 TreeNode* node; 17 nodeStack1.push(root); 18 while( !nodeStack1.empty()) 19 { 20 node = nodeStack1.top(); 21 nodeStack1.pop(); 22 nodeStack2.push(node); 23 if(node->left) 24 nodeStack1.push(node->left); 25 if(node->right) 26 nodeStack1.push(node->right); 27 } 28 29 while(!nodeStack2.empty()) 30 { 31 ans.push_back(nodeStack2.top()->val); 32 nodeStack2.pop(); 33 } 34 35 return ans; 36 } 37 };
优化:来源:
class Solution { public: vector<int> postorderTraversal(TreeNode* root) { vector<int> ans; if(root == NULL){ return ans; } stack<TreeNode*> q; TreeNode *p = root; TreeNode *pLast = NULL; while(p){ q.push(p); p = p->left; } while(!q.empty()){ p = q.top(); q.pop(); // 右无或者右看过,可以看中 if(!p->right || p->right==pLast){ ans.push_back(p->val); pLast = p; }else{ // 进入右 q.push(p); p = p->right; while(p){ q.push(p); p = p->left; } } } return ans; } };
标签:modal tor col 一个 || iss des else color
原文地址:https://www.cnblogs.com/jj81/p/11487437.html