leetcode-113-Path-Sum-II

描述


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,

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2
3
4
5
6
7
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1

Return:

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

分析


在一颗二叉树中找出所有节点值的和等于 sum 的路径。深度优先遍历。

解决方案1(Java)


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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private List<List<Integer>> result = new LinkedList<List<Integer>>();
public List<List<Integer>> pathSum(TreeNode root, int sum){
List<Integer> equalPath = new LinkedList<Integer>();
recursion(root, sum, equalPath);
return result;
}
public void recursion(TreeNode root, int sum, List<Integer> equalPath) {
if (root == null) {
return;
}
equalPath.add(new Integer(root.val));
if (root.val == sum && root.left == null && root.right == null) {
result.add(new LinkedList(equalPath));
} else {
recursion(root.left, sum-root.val, equalPath);
recursion(root.right, sum-root.val, equalPath);
}
equalPath.remove(equalPath.size()-1);
}
}

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