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leetcode_0112_Path_Sum.java
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// AC:
/**
* 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 solve(root, new LinkedList<>(), targetSum);
}
private boolean solve(TreeNode root, List<Integer> curPath, int target) {
if (root == null) {
return false;
}
curPath.add(root.val);
List<Integer> copy = new LinkedList<>(curPath);
// 叶子节点，进行判断
if (root.left == null && root.right == null) {
int tempSum = 0;
for (Integer i: curPath) {
tempSum += i;
}
if (tempSum == target) {
return true;
}
return false;
} else {
boolean checkLeft = solve(root.left, curPath, target);
if (checkLeft) {
return true;
}
// 这里很关键，要重置一次，否则 curPath 在上一步可能已经发生了变化
curPath = copy;
boolean checkRight = solve(root.right, curPath, target);
if (checkRight) {
return true;
}
return false;
}
}
}