113. Path Sum II-优快云博客

本文探讨了在给定二叉树和目标和的情况下，查找所有从根节点到叶子节点且路径上节点值之和等于给定目标和的路径问题。提供了三种不同语言（C++, Java, Python）的解决方案，包括递归、深度优先搜索和广度优先搜索算法实现。

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum 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  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

Approach #1: C++. [Recursive]

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        if (root == NULL) return ans;
        vector<int> temp;
        helper(root, temp, 0, sum);
        return ans;
    }
private:
    vector<vector<int>> ans;
    
    void helper(TreeNode* root, vector<int> temp, int cur, int sum) {  
        if (root == NULL) return;
        temp.push_back(root->val);
        if (root->left == NULL && root->right == NULL && cur+root->val == sum) {
            ans.push_back(temp);
            return;
        }
        helper(root->left, temp, cur+root->val, sum);
        helper(root->right, temp, cur+root->val, sum);
    }
};

Approach #2: Java.[DFS+Stack]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        int curSum = 0;
        TreeNode cur = root;
        TreeNode prev = null;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                curSum += cur.val;
                path.add(cur.val);
                cur = cur.left;
            }
            cur = stack.peek();
            if (cur.right != null & cur.right != prev) {
                cur = cur.right;
                continue;
            }
            if (cur.left == null && cur.right == null && curSum == sum)
                res.add(new ArrayList<Integer>(path));
            prev = cur;
            stack.pop();
            path.remove(path.size()-1);
            curSum -= cur.val;
            cur = null;
        }
        return res;
    }
}

Approach #3: Python. [BFS+Queue]

# 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 pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if root == None:
            return []
        
        res = []
        
        queue = [(root, root.val, [root.val])]
        
        while queue:
            curr, val, ls = queue.pop()
            
            if not curr.left and not curr.right and val == sum:
                res.append(ls)
            if curr.left:
                queue.append((curr.left, val+curr.left.val, ls+[curr.left.val]))
            if curr.right:
                queue.append((curr.right, val+curr.right.val, ls+[curr.right.val]))
                
        return res

转载于:https://www.cnblogs.com/ruruozhenhao/p/10093284.html