1. 二叉树的所有路径
深度优先遍历（DFS），使用临时变量（这里使用StringBuilder，也可以使用String，但是String是不可变对象，+= 操作将new出多个String对象，影响效率）保存当前节点的值，当节点的左右子树不都为null时，递归调用，当节点左右子树都为null时，即递归到叶子节点，将临时变量的值保存。LeetCode链接
public class LeetCode257 {
public static List<String> binaryTreePaths(TreeNode root) {
List<String> res = new ArrayList<>();
if (root == null) return res;
helper(root, new StringBuilder(), res);
return res;
}
private static void helper(TreeNode node, StringBuilder sb, List<String> res) {
sb.append(node.val);
if (node.left != null || node.right != null) {
sb.append("->");
if (node.left != null) {
helper(node.left, new StringBuilder(sb), res);
}
if (node.right != null) {
helper(node.right, new StringBuilder(sb), res);
}
}
if (node.left == null && node.right == null)
res.add(sb.toString());
}
public static void main(String[] args) {
TreeNode node = new TreeNode(1);
node.left = new TreeNode(2);
node.left.right = new TreeNode(5);
node.right = new TreeNode(3);
System.out.println(binaryTreePaths(node));
}
}
2. 路径总和
参考例子 1. 二叉树的所有路径，我们可以通过DFS遍历得到二叉树的所有路径。使用flag记录是否找到对应的分支，使用变量total累加每个节点的val值，当遍历的叶子节点时，判断total是否与给定的sum相等，相等则将flag置为true，提前退出递归。LeetCode链接
public class LeetCode112 {
private static boolean flag = false;
public static boolean hasPathSum(TreeNode root, int sum) {
if (root == null) return flag;
helper(root, sum, 0);
return flag;
}
private static void helper(TreeNode node, final int sum, int tmp) {
if (node == null || flag) return;
tmp += node.val;
if (node.left != null)
helper(node.left, sum, tmp);
if (node.right != null)
helper(node.right, sum, tmp);
if (node.left == null && node.right == null && tmp == sum) {
flag = true;
}
}
public static void main(String[] args) {
TreeNode node = new TreeNode(8);
node.left = new TreeNode(9);
node.right = new TreeNode(-6);
node.right.left = new TreeNode(5);
node.right.right = new TreeNode(9);
System.out.println(hasPathSum(node, 7));
}
}
3. 路径总和（2）
得到路径和等于给定值的所有路径。结合 1. 二叉树的所有路径 和 2. 路径总和，我们可以通过dfs获取所有路径，通过临时变量计算出当前路径总和是否等于sum值。多了一个使用List变量保存节点当前值的操作。LeetCode链接
public class LeetCode113 {
public static List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> res = new ArrayList<>();
helper(root, sum, 0, res, new ArrayList<>());
return res;
}
private static void helper(TreeNode node, int sum, int total, List<List<Integer>> res, List<Integer> tmpList) {
if (node == null) return;
total += node.val;
tmpList.add(node.val);
if (node.left != null)
helper(node.left, sum, total, res, new ArrayList<>(tmpList));
if (node.right != null)
helper(node.right, sum, total, res, new ArrayList<>(tmpList));
if (node.left == null && node.right==null && total == sum)
res.add(tmpList);
}
public static void main(String[] args) {
TreeNode node = new TreeNode(8);
node.left = new TreeNode(1);
node.right = new TreeNode(-3);
node.right.left = new TreeNode(4);
node.right.right = new TreeNode(3);
System.out.println(pathSum(node, 9));
}
}