[leetcode]二叉树遍历的非递归code-前中后+最近公共父节点

非递归框架

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def orderTraversal(self, root: TreeNode) -> List[int]:  
        p, r = root, None
        s = []
        ans = []
        while(len(s) > 0 or p is not None): # s模拟栈，p是当前节点
            if(p is not None):
               ...
            else:
                ...
        return ans

前序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        p, r = root, None
        s = []
        ans = []
        while(len(s) > 0 or p is not None):            
            if(p is not None):
                ans.append(p.val)
                s.append(p)
                p = p.left
            else:
                p = s[len(s)-1]    
                s.remove(s[len(s)-1])                
                p = p.right
        return ans

中序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:  
        p, r = root, None
        s = []
        ans = []
        while(len(s) > 0 or p is not None):            
            if(p is not None):
                s.append(p)
                p = p.left
            else:
                p = s[len(s)-1]    
                s.remove(s[len(s)-1])
                ans.append(p.val)
                p = p.right
        return ans

前序和中序只需交换一下append的位置即可，但后续可以吗？你猜呢

后序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:  
        p, r = root, None
        s = []
        ans = []
        while(len(s) > 0 or p is not None):            
            if(p is not None):
                s.append(p)
                p = p.left
            else:
                p = s[len(s)-1]
                # 如果没有右节点，或者已经走过右子树，可加入根节点
                if(p.right is None or p.right is r):
                    ans.append(p.val)
                    s.remove(s[len(s)-1])  
                    r = p # 记录根节点
                    p = None
                else:
                    p = p.right
        return ans

最近公共父节点

参考：https://blog.youkuaiyun.com/ACBattle/article/details/89482247