94_Binary Tree Inorder Traversal

94. Binary Tree Inorder Traversal

Level: medium

Tag: tree, stack, hash table

Question

Given a binary tree, return the inorder traversal of its nodes' values.

Example 1

Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2]

Solution 1: recursive

# 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]
        """
        ret = []
        self.dps(root, ret)
        return ret


    def dps(self, root, ret):
        """
        :type root: TreeNode
        :     ret: List[int]
        :rtype List[int]
        """
        if root:
            self.dps(root.left, ret)
            ret.append(root.val)
            self.dps(root.right, ret)

Solution 2: iteratively (my solution)

use stack to store right child, node value, and left child repeatedly. When node value is popped out, add to result, otherwise repeat.

# 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]
        """
        stack = [root]
        ret = []
        while stack:
            node = stack.pop()
            if type(node) == int:
                ret.append(node)
            elif node:         
                stack.append(node.right)
                stack.append(node.val)
                stack.append(node.left)
        return ret

Idea: 假设树为:

                1

               /   \

               2    3

               /   \  /   \

               4     5  6    7

我们使用一个栈来解决问题。步骤如下:

一,我们将根节点1入栈,如果有左孩子,依次入栈,那么入栈顺序为:1,2,4。由于4的左子树为空,停止入栈,此时栈为{1,2,4}。

二,此时将4出栈,并遍历4,由于4也没有右孩子,那么根据中序遍历的规则,我们显然应该继续遍历4的父亲2,情况是这样。所以我们继续将2出栈并遍历2,2存在右孩子,将5入栈,此时栈为{1,5}。

三,5没有孩子,则将5出栈并遍历5,这也符合中序遍历的规则。此时栈为{1}。

四,1有右孩子,则将1出栈并遍历1,然后将右孩子3入栈,并继续以上三个步骤即可。

栈的变化过程:{1}->{1,2}->{1,2,4}->{1,2}->{1}->{1,5}->{1}->{}->{3}->{3,6}->{3}->{}->{7}->{}。

# 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]
        """
        stack = []
        ret = []
        node = root
        while stack or node:
            while node:
                stack.append(node)
                node = node.left
            node = stack.pop()
            ret.append(node.val)
            node = node.right
        return ret

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