LeetCode解題之Binary Tree Inorder Traversal

原題

不用遞歸來實現樹的中序遍歷。

注意點：

例子：

輸入: {1,#,2,3}

   1
    \
     2
    /
   3

輸出: [1,3,2]

解題思路

通過棧來實現，從根節點開始，不斷尋找左節點，並把這些節點依次壓入棧內，只有在該節點沒有左節點或者它的左子樹都已經遍歷完成後，它才會從棧內彈出，這時候訪問該節點，並它的右節點當做新的根節點一樣不斷遍歷。

AC源碼

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


class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result = []
        stack = []
        p = root
        while p or stack:
            # Save the nodes which have left child
            while p:
                stack.append(p)
                p = p.left
            if stack:
                p = stack.pop()
                # Visit the middle node
                result.append(p.val)
                # Visit the right subtree
                p = p.right
        return result


if __name__ == "__main__":
    n1 = TreeNode(1)
    n2 = TreeNode(2)
    n3 = TreeNode(3)
    n1.right = n2
    n2.left = n3
    assert Solution().inorderTraversal(n1) == [1, 3, 2]