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Still 3 methods:
Same with inorder, but post order is a little different. We need to treat it as reverse result of preorder.
So we go to right branch first, then go back to left branch
1. recursive (use memory stack);
1 /** 2 * Definition for binary tree 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 void getTree(TreeNode *root, vector<int> &result) { 13 if (!root) return; 14 getTree(root->left, result); 15 getTree(root->right, result); 16 result.push_back(root->val); 17 } 18 vector<int> postorderTraversal(TreeNode *root) { 19 vector<int> result; 20 getTree(root, result); 21 return result; 22 } 23 }; 24
2. directly use stack:
1 /** 2 * Definition for binary tree 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> result; 14 stack<TreeNode *> s; 15 if (!root) return result; 16 while (root || !s.empty()) { 17 if (root) { 18 result.push_back(root->val); 19 s.push(root->left); 20 root = root->right; 21 } else { 22 root = s.top(); 23 s.pop(); 24 } 25 } 26 for (int i = 0; i < result.size()/2; i++) { 27 [](int &a, int &b) {int t = a; a = b; b = t;}(result[i], result[result.size()-i-1]); 28 } 29 return result; 30 } 31 }; 32
3. two pointers:
1 /** 2 * Definition for binary tree 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> result; 14 if (!root) return result; 15 TreeNode *prev = NULL; 16 while (root) { 17 result.push_back(root->val); 18 if (!root->right) { 19 root = root->left; 20 } else { 21 prev = root->right; 22 while(prev->left && prev->left != root) prev = prev->left; 23 if (!prev->left) { 24 prev->left = root; 25 root = root->right; 26 } else { 27 prev->left = NULL; 28 root = root->left; 29 result.pop_back(); 30 } 31 } 32 } 33 for (int i = 0; i < result.size()/2; i++) { 34 [](int &a, int &b) {int t = a; a = b; b = t;}(result[i], result[result.size()-i-1]); 35 } 36 return result; 37 } 38 }; 39
LeetCode – Refresh – Binary Tree Post Order Traversal
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原文地址:http://www.cnblogs.com/shuashuashua/p/4346191.html