【LeetCode】Path Sum II 解题报告

标签：dfs   递归   二叉树   leetcode   

【题目】

Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.

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

return

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

【解析】

DFS，递归实现。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        if (root == null) return res;
        List<Integer> list = new ArrayList<Integer>();
        list.add(root.val);
        dfs(root, sum - root.val, list);
        return res;
    }
    
    public void dfs(TreeNode root, int sum, List<Integer> list) {
        
        if (root.left == null && root.right == null && sum == 0) {
            res.add(list);
            return;
        }
        if (root.left != null) {
            List<Integer> leftList = new ArrayList<Integer>(list);
            leftList.add(root.left.val);
            dfs(root.left, sum - root.left.val, leftList);
        }
        if (root.right != null) {
            List<Integer> rightList = new ArrayList<Integer>(list);
            rightList.add(root.right.val);
            dfs(root.right, sum - root.right.val, rightList);
        }
    }
}

【LeetCode】Path Sum II 解题报告

标签：dfs   递归   二叉树   leetcode   

原文地址：http://blog.csdn.net/ljiabin/article/details/41412579