題目

https://leetcode.com/problems/path-sum/

https://leetcode.com/problems/path-sum-ii/

題解

簡單的遍歷二叉樹，不解釋，直接上代碼

112. Path Sum

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/ class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {return dfs(root, 0, targetSum);}public boolean dfs(TreeNode node, int sum, int target) {if (node == null) return false;if (node.left == null && node.right == null) return sum + node.val == target;return dfs(node.left, sum + node.val, target) || dfs(node.right, sum + node.val, target);} }

113. Path Sum II

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/ class Solution {List<List<Integer>> res;public List<List<Integer>> pathSum(TreeNode root, int targetSum) {res = new ArrayList<>();dfs(root, 0, targetSum,new ArrayList<>());return res;}public void dfs(TreeNode node, int sum, int target, List<Integer> list) {if (node == null) return;List<Integer> newList = new ArrayList<>(list);newList.add(node.val);if (node.left == null && node.right == null && sum + node.val == target) { // 是葉子res.add(newList);}dfs(node.left, sum + node.val, target, newList);dfs(node.right, sum + node.val, target, newList);} }

總結(jié)

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