[LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal
Pre: node 先, Inorder: node in, Postorder: node 最后
PreOrder | Inorder | PostOrder |
node-> left -> right | left -> node ->right | left -> right ->node |
Recursive method
实际上代码是一样, 就是把ans.append(root.val) 放在如上表先, 中, 后就是pre, in, post order了.
1) PreOrder traversal
ans = []
def preOrder(self, root):
if not root: return
ans.append(root.val)
preOrder(root.left)
preOrder(root.right) preOrder(root)
return ans
2) Inorder traversal Worst S: O(n), average is O(lgn)
ans = []
def inOrder(self, root):
if not root: return
inOrder(root.left)
ans.append(root.val)
inOrder(root.right) inOrder(root)
return ans
3) PostOrder traversal
ans = []
def postOrder(self, root):
if not root: return
postOrder(root.left)
postOrder(root.right)
ans.append(root.val)
postOrder(root)
return ans
Iterable method
1) Preorder traversal --- Just use stack.
node-> left -> right
def Preorder(self, root):
if not root: return []
ans, stack = [], [root]
while stack:
node = stack.pop()
ans.append(node.val)
if node.right:
stack.append(node.right)
if node.left:
stack.append(node.left)
return ans
2) inOrder traversal
left -> node ->right
def inOrder(self, root):
ans, stack = [], []
while True:
while root:
stack.append(root)
root = root.left
if not stack: return ans
node = stack.pop()
ans.append(node.val)
root = node.right
seconde inOrder traversal
def inOrder(self, root):
ans, stack = [], []
while stack or root:
if root:
stack.append(root)
root = root.left
else:
node = stack.pop()
ans.append(node.val)
root = node.right
return ans
3) PostOrder traversal
left -> right ->node
由于我们已经知道如何用preorder, 所以我们知道用 node-> left -> right, 所以我们可以用类似于preorder的做法, 将node-> right -> left 做出来, 最后返回reverse 的ans即可.
def PostOrder(self, root):
if not root: return []
stack, ans = [root], []
while stack:
node = stack.pop()
ans.append(node.val)
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return ans[::-1]
second PostOrder traversal
利用hash table, 但是我们直接将这个hash table append进入stack, 跟node形成一个tuple.
def PostOrder(self, root):
if not root: return []
stack, ans = [(root, False)], []
while stack:
node, visited = stack.pop()
if visited:
ans.append(node.val)
else:
stack.append((node, True))
if node.right:
stack.append((node.right, False))
if node.left:
stack.append((node.left, False))
return ans
Questions:
[LeetCode]94, 144, 145 Binary Tree InOrder, PreOrder, PostOrder Traversal_Medium
[LeetCode] 589. N-ary Tree Preorder Traversal_Easy
[LeetCode] 590. N-ary Tree Postorder Traversal_Easy
[LeetCode] 98. Validate Binary Search Tree_Medium
[LeetCode] 230. Kth Smallest Element in a BST_Medium tag: Inorder Traversal
[LeetCode] 285. Inorder Successor in BST_Medium tag: Inorder Traversal
[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal_Medium tag: Tree Traversal
[LeetCode] 255. Verify Preorder Sequence in Binary Search Tree_Medium tag: Preorder Traversal, tree
[LeetCode] 331. Verify Preorder Serialization of a Binary Tree_Medium tag: stack
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