Leetcode 112.路径总和

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112.路径总和

题目

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。

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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。

方法

方法1：递归深度优先搜索（深度优先遍历）（先序遍历）

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# 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 not root:
            return False
        if not root.left and not root.right and root.val==sum:
            return True

        sum-=root.val
        return self.hasPathSum(root.left,sum) or self.hasPathSum(root.right,sum)

方法2：非递归深度优先搜索（深度优先遍历）（先序遍历）

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# 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 not root:
            return False

        stack=[(root,sum)]
        while stack:
            node,sum=stack.pop()
            if not node.left and not node.right and node.val==sum:
                return True
            if node.left:
                stack.append((node.left,sum-node.val))
            if node.right:
                stack.append((node.right,sum-node.val))
        return False
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# 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 == None:
            return False
        if root.val == sum and not root.left and not root.right:
            return True
        l = self.hasPathSum(root.left, sum-root.val)
        if l:
            return True
        r = self.hasPathSum(root.right, sum-root.val)
        if r:
            return True
        return False