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] 106. Construct Binary Tree from Postorder 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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