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?
Analyse: root->left, root->right, root.
1. Recursion
Runtime: 0ms.
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> result; 14 if(!root) return result; 15 16 postorder(root, result); 17 return result; 18 } 19 void postorder(TreeNode* root, vector<int>& result){ 20 if(root->left) postorder(root->left, result); 21 if(root->right) postorder(root->right, result); 22 result.push_back(root->val); 23 } 24 };
2. Iteration: Only when the children of the parent are visited or the parent has no children, this parent can be pushed into the vector as visited. Otherwise, we should push its right child and left child respectively. See reference from Binary Tree Postorder Traversal
Runtime: 0ms.
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> result; 14 if(!root) return result; 15 stack<TreeNode* >stk; 16 stk.push(root); 17 TreeNode* visited = root; 18 19 while(!stk.empty()){ 20 root = stk.top(); 21 22 if(root->right == visited || root->left == visited || (!root->left) && (!root->right)){ 23 result.push_back(root->val); 24 stk.pop(); 25 visited = root; 26 } 27 else{ 28 if(root->right) stk.push(root->right); 29 if(root->left) stk.push(root->left); 30 } 31 } 32 return result; 33 } 34 };