112. 路径总和

给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。

说明: 叶子节点是指没有子节点的节点。

示例: 
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

思路：利用递归思想，sum - root.val的值是否为零，不为零就进行递归运算，为零判断是否为独立的根节点。

代码：

# 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 hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root is None:
            return False
        sum = sum - root.val
        if sum == 0:
            if root.left is None and root.right is None:
                return True
        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)
    

二叉树的三种（递归）遍历方式

二叉树先序遍历

代码如下：

# 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 preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return [] 
        path = []
        def order(root):
            path.append(root.val)
            if root.left is not None:
                order(root.left)
            if root.right is not None:
                order(root.right)
        order(root)
        return path
                

二叉树中序遍历

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代码如下：

# 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]
        """
        if root is None:
            return []
        path = []
        def order(root):
            if root.left is not None:
                order(root.left)
            path.append(root.val)
            if root.right is not None:
                order(root.right)
        order(root)
        return path
    

二叉树的后序遍历

代码如下：

# 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 postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        path = []
        def order(root):
            if root.left is not None:
                order(root.left)
            if root.right is not None:
                order(root.right)
            path.append(root.val)
        order(root)
        return path