Path Sum II

标签：二叉树

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    2  5   1

return

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

#include<iostream>
#include<vector>
using namespace std;

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

void getPath(vector<vector<int> > &ResultVector, vector<int> &Path, TreeNode *curnode, int sum)
{
	if (curnode == NULL)
		return;
	else if (!curnode->left&&!curnode->right){
		if (sum-curnode->val==0){
			ResultVector.push_back(Path);
			Path.erase(Path.end() - 1);
			return;
		}
	}
	Path.push_back(curnode->val);
	getPath(ResultVector, Path, curnode->left, sum - curnode->val);
	getPath(ResultVector, Path, curnode->right, sum - curnode->val);
	Path.erase(Path.end() - 1);
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
	vector<vector<int> > ResultVector;
	vector<int> Path;
	getPath(ResultVector, Path, root, sum);
	return ResultVector;
}

Path Sum II

标签：二叉树

原文地址：http://blog.csdn.net/li_chihang/article/details/44749383