[Leetcode] Path Sum I,II,III

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112. Path Sum I: 点击打开链接

/** * Definition for a binary tree node.         //判断是否有这样的路径满足和为sum * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        return helper(root,sum);    }        public boolean helper(TreeNode root,int sum){        if(root==null){            return false;        }                if(root.left==null && root.right==null){            if(root.val==sum){                return true;            }        }        return helper(root.left,sum-root.val) || helper(root.right,sum-root.val);    }}

113. Path Sum II:点击打开链接

/** * Definition for a binary tree node.         //写出从根节点开始的和为sum的每一条路径 * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> pathSum(TreeNode root, int sum) {        List<List<Integer>> result=new ArrayList<>();        List<Integer> path=new ArrayList<>();        helper(root,sum,result,path);        return result;    }        public void helper(TreeNode root,int target,List<List<Integer>> result,List<Integer> path){        if(root==null){            return;        }                if(root.left==null && root.right==null){            if(root.val==target){                path.add(root.val);                result.add(path);                return;            }        }                if(root.left!=null){            List<Integer> left=new ArrayList<>(path);     //下一层的添加在之前的基础上添加，因此deep copy            left.add(root.val);                                        helper(root.left,target-root.val,result,left);        }                if(root.right!=null){            List<Integer> right=new ArrayList<>(path);            right.add(root.val);            helper(root.right,target-root.val,result,right);        }    }}

437. Path Sum III:点击打开链接

/** * Definition for a binary tree node.           //数出从任意节点开始的和为sum的路径的条数 * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int pathSum(TreeNode root, int sum) {        //根节点开始的满足条件的路径和 + 左子树的情况 + 右子树的情况        if (root == null) return 0;        return helper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);    }        private int helper(TreeNode node, int sum) {        //从任意node开始满足条件的路径和        if (node == null) return 0;        int left=helper(node.left, sum - node.val);        int right=helper(node.right, sum - node.val);        return (node.val == sum ? 1 : 0) +left+right;    }}

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