C++实现LeetCode(112.二叉树的路径和)

[LeetCode] 112. Path Sum 二叉树的路径和

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

这道题给了一棵二叉树，问是否存在一条从跟结点到叶结点到路径，使得经过到结点值之和为一个给定的 sum 值，这里需要用深度优先算法 DFS 的思想来遍历每一条完整的路径，也就是利用递归不停找子结点的左右子结点，而调用递归函数的参数只有当前结点和 sum 值。首先，如果输入的是一个空结点，则直接返回 false，如果如果输入的只有一个根结点，则比较当前根结点的值和参数 sum 值是否相同，若相同，返回 true，否则 false。 这个条件也是递归的终止条件。下面就要开始递归了，由于函数的返回值是 Ture/False，可以同时两个方向一起递归，中间用或 || 连接，只要有一个是 True，整个结果就是 True。递归左右结点时，这时候的 sum 值应该是原 sum 值减去当前结点的值，参见代码如下：

解法一：

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) return false;
        if (!root->left && !root->right && root->val == sum ) return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};

我们也可以使用迭代的写法，这里用的也是先序遍历的迭代写法，先序遍历二叉树，左右子结点都需要加上其父结点值，这样当遍历到叶结点时，如果和 sum 相等了，那么就说明一定有一条从 root 过来的路径。注意这里不必一定要先处理右子结点，调换下顺序也是可以的，因为不论是先序遍历的根-左-右，还是根-右-左，并不会影响到找路径，参见代码如下：

解法二：

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) return false;
        stack<TreeNode*> st{{root}};
        while (!st.empty()) {
            TreeNode *t = st.top(); st.pop();
            if (!t->left && !t->right) {
                if (t->val == sum) return true;
            }
            if (t->right) {
                t->right->val += t->val;
                st.push(t->right);
            }
            if (t->left) {
                t->left->val += t->val;
                st.push(t->left);
            }
        }
        return false;
    }
};

到此这篇关于C++实现LeetCode(112.二叉树的路径和)的文章就介绍到这了,更多相关C++实现二叉树的路径和内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(113.二叉树路径之和之二)

    [LeetCode] 113. Path Sum II 二叉树路径之和之二 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22,  5 / \ 4   8 /      / \ 11  13  4 /  \         / \ 7  

  • C++实现LeetCode(51.N皇后问题)

    C++实现LeetCode(51.N皇后问题)

    [LeetCode] 51. N-Queens N皇后问题 The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a di

  • C++实现LeetCode(46.全排列)

    [LeetCode] 46. Permutations 全排列 Given a collection of distinct integers, return all possible permutations. Example: Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 这道题是求全排列问题,给的输入数组没有重复项,这跟之前的那道 Combinations 和类似,解法基本相同

  • C++实现LeetCode(39.组合之和)

    [LeetCode] 39. Combination Sum 组合之和 Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may b

  • C++实现LeetCode(144.二叉树的先序遍历)

    [LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历 Given a binary tree, return the preorder traversal of its nodes' values. Example: Input:  [1,null,2,3] 1 \ 2 / 3 Output:  [1,2,3] Follow up: Recursive solution is trivial, could you do it iterat

  • C++实现LeetCode(90.子集合之二)

    [LeetCode] 90. Subsets II 子集合之二 Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example

  • C++实现LeetCode(78.子集合)

    [LeetCode] 78. Subsets 子集合 Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example, If S = [1,2,3], a solution is:

  • C++实现LeetCode(38.计数和读法)

    [LeetCode] 38. Count and Say 计数和读法 The count-and-say sequence is the sequence of integers with the first five terms as following: 1.     1 2.     11 3.     21 4.     1211 5.     111221 1 is read off as "one 1" or 11. 11 is read off as "two

  • C++实现LeetCode(112.二叉树的路径和)

    [LeetCode] 112. Path Sum 二叉树的路径和 Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children. Example: Given the below bi

  • java编程求二叉树最大路径问题代码分析

    题目: Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 节点可能为负数,寻找一条最路径使得所经过节点和最大.路径可以开始和结束于任何节点但是不能走回头路. 这道题虽然

  • Java实现打印二叉树所有路径的方法

    本文实例讲述了Java实现打印二叉树所有路径的方法.分享给大家供大家参考,具体如下: 问题: 给一个二叉树,把所有的路径都打印出来. 比如,对于下面这个二叉树,它所有的路径为: 8 -> 3 -> 1 8 -> 2 -> 6 -> 4 8 -> 3 -> 6 -> 7 8 -> 10 -> 14 -> 13 思路: 从根节点开始,把自己的值放在一个数组里,然后把这个数组传给它的子节点,子节点同样把自己的值放在这个数组里,又传给自己的子节点,

  • C++实现LeetCode(94.二叉树的中序遍历)

    [LeetCode] 94. Binary Tree Inorder Traversal 二叉树的中序遍历 Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively

  • C++实现LeetCode(63.不同的路径之二)

    C++实现LeetCode(63.不同的路径之二)

    [LeetCode] 63. Unique Paths II 不同的路径之二 A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right c

  • C++实现LeetCode(62.不同的路径)

    C++实现LeetCode(62.不同的路径)

    [LeetCode] 62. Unique Paths 不同的路径 A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner

  • C++实现LeetCode(145.二叉树的后序遍历)

    [LeetCode] 145. Binary Tree Postorder Traversal 二叉树的后序遍历 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3},    1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you d

  • C++实现LeetCode(104.二叉树的最大深度)

    [LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度 Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Note: A leaf is a node with no child

随机推荐