LeetCode解题之Binary Tree Inorder Traversal

原题

不用递归来实现树的中序遍历。

注意点：

• 无

例子：

输入: {1,#,2,3}

   1
         2
    /
   3

输出: [1,3,2]

解题思路

通过栈来实现，从根节点开始，不断寻找左节点，并把这些节点依次压入栈内，只有在该节点没有左节点或者它的左子树都已经遍历完成后，它才会从栈内弹出，这时候访问该节点，并它的右节点当做新的根节点一样不断遍历。

AC源码

# 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]

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