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This is a fundamental and yet classic problem. I share my three solutions here:
O(n)
time and O(n)
space;O(n)
time and O(n)
space (considering the costs of call stack);O(n)
time and O(1)
space!!!Iterative solution using stack:
1 vector<int> postorderTraversal(TreeNode* root) { 2 vector<int> nodes; 3 stack<TreeNode*> toVisit; 4 TreeNode* curNode = root; 5 TreeNode* lastNode = NULL; 6 while (curNode || !toVisit.empty()) { 7 if (curNode) { 8 toVisit.push(curNode); 9 curNode = curNode -> left; 10 } 11 else { 12 TreeNode* topNode = toVisit.top(); 13 if (topNode -> right && lastNode != topNode -> right) 14 curNode = topNode -> right; 15 else { 16 nodes.push_back(topNode -> val); 17 lastNode = topNode; 18 toVisit.pop(); 19 } 20 } 21 } 22 return nodes; 23 }
Recursive solution:
1 void postorder(TreeNode* root, vector<int>& nodes) { 2 if (!root) return; 3 postorder(root -> left, nodes); 4 postorder(root -> right, nodes); 5 nodes.push_back(root -> val); 6 } 7 vector<int> postorderTraversal(TreeNode* root) { 8 vector<int> nodes; 9 postorder(root, nodes); 10 return nodes; 11 }
Morris traversal:
1 void reverseNodes(TreeNode* start, TreeNode* end) { 2 if (start == end) return; 3 TreeNode* x = start; 4 TreeNode* y = start -> right; 5 TreeNode* z; 6 while (x != end) { 7 z = y -> right; 8 y -> right = x; 9 x = y; 10 y = z; 11 } 12 } 13 void reverseAddNodes(TreeNode* start, TreeNode* end, vector<int>& nodes) { 14 reverseNodes(start, end); 15 TreeNode* node = end; 16 while (true) { 17 nodes.push_back(node -> val); 18 if (node == start) break; 19 node = node -> right; 20 } 21 reverseNodes(end, start); 22 } 23 vector<int> postorderTraversal(TreeNode* root) { 24 vector<int> nodes; 25 TreeNode* dump = new TreeNode(0); 26 dump -> left = root; 27 TreeNode* curNode = dump; 28 while (curNode) { 29 if (curNode -> left) { 30 TreeNode* predecessor = curNode -> left; 31 while (predecessor -> right && predecessor -> right != curNode) 32 predecessor = predecessor -> right; 33 if (!(predecessor -> right)) { 34 predecessor -> right = curNode; 35 curNode = curNode -> left; 36 } 37 else { 38 reverseAddNodes(curNode -> left, predecessor, nodes); 39 predecessor -> right = NULL; 40 curNode = curNode -> right; 41 } 42 } 43 else curNode = curNode -> right; 44 } 45 return nodes; 46 }
[LeetCode] Binary Tree Postorder Traversal
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原文地址:http://www.cnblogs.com/jcliBlogger/p/4548024.html