There are three types of depth-first traversal: pre-order,in-order, and post-order.
For a binary tree, they are defined as operations recursively at each node, starting with the root node as follows:
Pre-order
Visit the root.
Traverse the left subtree.
Traverse the right subtree.
iterativePreorder(node)
parentStack = empty stack
parentStack.push(null)
top = node
while ( top != null )
visit( top )
if ( top.right != null )
parentStack.push(top.right)
if ( top.left != null )
parentStack.push(top.left)
top = parentStack.pop()
In-order
Traverse the left subtree.
Visit root.
Traverse the right subtree.
iterativeInorder(node)
parentStack = empty stack
while (not parentStack.isEmpty() or node ≠ null)
if (node ≠ null)
parentStack.push(node)
node = node.left
else
node = parentStack.pop()
visit(node)
node = node.right
Post-order
Traverse the left subtree.
Traverse the right subtree.
Visit the root.
iterativePostorder(node)
parentStack = empty stack
lastnodevisited = null
while (not parentStack.isEmpty() or node ≠ null)
if (node ≠ null)
parentStack.push(node)
node = node.left
else
peeknode = parentStack.peek()
if (peeknode.right ≠ null and lastnodevisited ≠ peeknode.right)
/* if right child exists AND traversing node from left child, move right */
node = peeknode.right
else
parentStack.pop()
visit(peeknode)
lastnodevisited = peeknode
Morris Travel
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> inorderTraversal(TreeNode *root) { 13 vector<int> ret; 14 if (!root) return ret; 15 TreeNode *cur = root; 16 17 while (cur) { 18 if (!cur->left) { 19 ret.push_back(cur->val); 20 cur = cur->right; 21 } else { 22 TreeNode *rightmost = cur->left; 23 while (rightmost->right != NULL && rightmost->right != cur) rightmost = rightmost->right; 24 if (rightmost->right == cur) { 25 rightmost->right = NULL; 26 ret.push_back(cur->val); 27 cur = cur->right; 28 } else { 29 rightmost->right = cur; 30 cur = cur->left; 31 } 32 } 33 } 34 35 return ret; 36 } 37 };
后序：
