Binary Tree Paths | LeetCode OJ
Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
https://leetcode.com/problems/binary-tree-paths/discuss/68282/Clean-Java-solution-(Accepted)-without-any-helper-recursive-function
https://leetcode.com/articles/binary-tree-paths/
Method 3: use the StringBuilder, no recursion (idea from Raina)
This method get the list of Nodes first and then get the corresponding string, and add the strings to a list one by one.
Method 2: use the StringBuilder, recursion
This creates too many stringbuilder.
http://sbzhouhao.net/LeetCode/LeetCode-Binary-Tree-Paths.html
This can avoid String concatenation.
https://leetcode.com/discuss/55451/clean-solution-accepted-without-helper-recursive-function
List<String> paths = new LinkedList<>();
if(root == null) return paths;
if(root.left == null && root.right == null){
paths.add(root.val+"");
return paths;
}
for (String path : binaryTreePaths(root.left)) {
paths.add(root.val + "->" + path);
}
for (String path : binaryTreePaths(root.right)) {
paths.add(root.val + "->" + path);
}
return paths;
}
https://xuezhashuati.blogspot.com/2016/02/binary-tree-paths.html
https://zhengyang2015.gitbooks.io/lintcode/content/binary_tree_paths_480.html
Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
All root-to-leaf paths are:1 / \ 2 3 \ 5
["1->2->5", "1->3"]
https://segmentfault.com/a/1190000003465753 Use StringBuilder
List<String> res = new ArrayList<String>();
public List<String> binaryTreePaths(TreeNode root) {
if(root != null) findPaths(root,String.valueOf(root.val));
return res;
}
private void findPaths(TreeNode n, String path){
if(n.left == null && n.right == null) res.add(path);
if(n.left != null) findPaths(n.left, path+"->"+n.left.val);
if(n.right != null) findPaths(n.right, path+"->"+n.right.val);
}
class Solution:
def binaryTreePaths(self, root):
if not root:
return []
return [str(root.val) + '->' + path
for kid in (root.left, root.right) if kid
for path in self.binaryTreePaths(kid)] or [str(root.val)]
public void construct_paths(TreeNode root, String path, LinkedList<String> paths) {
if (root != null) {
path += Integer.toString(root.val);
if ((root.left == null) && (root.right == null)) // if reach a leaf
paths.add(path); // update paths
else {
path += "->"; // extend the current path
construct_paths(root.left, path, paths);
construct_paths(root.right, path, paths);
}
}
}
public List<String> binaryTreePaths(TreeNode root) {
LinkedList<String> paths = new LinkedList();
construct_paths(root, "", paths);
return paths;
}
https://leetcode.com/problems/binary-tree-paths/discuss/68282/Clean-Java-solution-(Accepted)-without-any-helper-recursive-function
Lot of recursive solutions on this forum involves creating a helper recursive function with added parameters. The added parameter which usually is of the type List , carries the supplementary path information. However, the approach below doesn't use such a helper function.
public List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new LinkedList<>();
if(root == null) return paths;
if(root.left == null && root.right == null){
paths.add(root.val+"");
return paths;
}
for (String path : binaryTreePaths(root.left)) {
paths.add(root.val + "->" + path);
}
for (String path : binaryTreePaths(root.right)) {
paths.add(root.val + "->" + path);
}
return paths;
}https://leetcode.com/articles/binary-tree-paths/
public List<String> binaryTreePaths(TreeNode root) {
LinkedList<String> paths = new LinkedList();
if (root == null)
return paths;
LinkedList<TreeNode> node_stack = new LinkedList();
LinkedList<String> path_stack = new LinkedList();
node_stack.add(root);
path_stack.add(Integer.toString(root.val));
TreeNode node;
String path;
while (!node_stack.isEmpty()) {
node = node_stack.pollLast();
path = path_stack.pollLast();
if ((node.left == null) && (node.right == null))
paths.add(path);
if (node.left != null) {
node_stack.add(node.left);
path_stack.add(path + "->" + Integer.toString(node.left.val));
}
if (node.right != null) {
node_stack.add(node.right);
path_stack.add(path + "->" + Integer.toString(node.right.val));
}
}
return paths;
}
http://yueguo1217.com/leetcode-binary-tree-paths-easy-8-in-java/Method 3: use the StringBuilder, no recursion (idea from Raina)
This method get the list of Nodes first and then get the corresponding string, and add the strings to a list one by one.
public String getPath(List Nodes) {
StringBuilder sB = new StringBuilder();
sB.append(Nodes.get(0).val);
if(Nodes.size()!=1) {
for(int i=1; i<Nodes.size(); i++)
sB.append("->").append(Nodes.get(i).val);
}
return sB.toString();
}
public List binaryTreePaths(TreeNode root) {
List thePath = new ArrayList();
if(root!= null) {
List theStack = new ArrayList();
theStack.add(root);
TreeNode preNode = null;
// the process
while(!theStack.isEmpty()) {
TreeNode peekNode = theStack.get(theStack.size()-1);
// a peek Node for both downward and upward look
/* downward
(judge by relations b/w pre and peek Nodes)*/
if((preNode==null)||(preNode.left==peekNode)||
(preNode.right==peekNode)) {
if(peekNode.left!=null)
theStack.add(peekNode.left);
else if(peekNode.right!=null)
theStack.add(peekNode.right);
else { // the final part
thePath.add(this.getPath(theStack));//this
}
}
// upward (judge by relations b/w pre and peek Nodes)
// if move backward to search for right branch path,
the peek becomes the pre's parent
else if((peekNode.left==preNode)&&
(peekNode.right!=null)) {
theStack.add(peekNode.right);
/* once a new branch path begins, the peek
becomes the pre's child again
and the focus changes to this new pair */
} else theStack.remove(peekNode); // key line
preNode = peekNode; // key line
}
}
return thePath;
}
Method 2: use the StringBuilder, recursion
This creates too many stringbuilder.
public List<String> binaryTreePaths(TreeNode root) {
//initialize the path
List<String> thePath = new ArrayList<String> ();
if(root==null) return thePath;
recordPath(root, new StringBuilder(), thePath);
return thePath;
}
public void recordPath(TreeNode theRoot, StringBuilder sB,
List<String> path) {
/* StringBuilder is handy when "+" types of v.r.s and
strings together */
//the final part for all (add the string to a list)
if((theRoot.left==null)&&(theRoot.right==null)) {
sB.append(theRoot.val);
path.add(sB.toString());
return; //back to the (method) caller
}
sB.append(theRoot.val);
sB.append("->");
//get the duplicates of parents automatically
if(theRoot.left!=null)
recordPath(theRoot.left, new StringBuilder(sB), path); // A
if(theRoot.right!=null)
recordPath(theRoot.right, new StringBuilder(sB), path);
/* we use NEW StringBuilder to avoid add vals of the right
directly to the results with the val of left when theRoot
has two children, coz the sB changed after the lineA */
}
because when the root has two children, it should comes out two StringBuilder all base on the original root StringBuilder, for convenience, we just create a new StringBuilder when root has a child.http://sbzhouhao.net/LeetCode/LeetCode-Binary-Tree-Paths.html
This can avoid String concatenation.
public List<String> binaryTreePaths(TreeNode root) { List<String> result = new ArrayList<>(); if (root == null) { return result; } helper(result, new ArrayList<>(), root); return result; } private void helper(List<String> result, List<Integer> crt, TreeNode root) { if (root == null) { return; } else { crt.add(root.val); } if (root.right == null && root.left == null) { result.add(getString(crt)); return; } if (root.left != null) { helper(result, new ArrayList<>(crt), root.left); } if (root.right != null) { helper(result, new ArrayList<>(crt), root.right); } } private String getString(List<Integer> crt) { if (crt == null || crt.isEmpty()) { return null; } StringBuilder sb = new StringBuilder(); sb.append(crt.get(0)); for (int i = 1; i < crt.size(); i++) { sb.append("->").append(crt.get(i)); } return sb.toString(); }http://www.chenguanghe.com/binary-tree-paths/
public List<String> binaryTreePaths(TreeNode root) {
List<String> res = new ArrayList<String>();
if(root == null)
return res;
if(root.left == null && root.right == null)
res.add(root.val + "");
else{
if(root.left != null)
res.addAll(binaryTreePaths(root.left));
if(root.right != null)
res.addAll(binaryTreePaths(root.right));
for(int i = 0 ; i < res.size(); i++)
res.set(i,root.val + "->" + res.get(i));
}
return res;
}
http://blog.welkinlan.com/2015/09/07/binary-tree-paths-leetcode-java/
public List<String> binaryTreePaths(TreeNode root) {
List<String> result = new ArrayList<String>();
if (root == null) {
return result;
}
findPaths(root, result, root.val + "");
return result;
}
private void findPaths(TreeNode root, List<String> result, String cur) {
if (root.left == null && root.right == null) {
result.add(cur);
return;
}
if (root.left != null) {
findPaths(root.left, result, cur + "->" + root.left.val);
}
if (root.right != null) {
findPaths(root.right, result, cur + "->" + root.right.val);
}
}
http://www.geeksforgeeks.org/given-a-binary-tree-print-all-root-to-leaf-paths//* Recursive helper function -- given a node, and an array containing the path from the root node up to but not including this node, print out all the root-leaf paths.*/void printPathsRecur(struct node* node, int path[], int pathLen) { if (node==NULL) return; /* append this node to the path array */ path[pathLen] = node->data; pathLen++; /* it's a leaf, so print the path that led to here */ if (node->left==NULL && node->right==NULL) { printArray(path, pathLen); } else { /* otherwise try both subtrees */ printPathsRecur(node->left, path, pathLen); printPathsRecur(node->right, path, pathLen); }}
smart one. Too many array copying makes it slower when compared with other answers.
public List<String> binaryTreePaths(TreeNode root) {List<String> paths = new LinkedList<>();
if(root == null) return paths;
if(root.left == null && root.right == null){
paths.add(root.val+"");
return paths;
}
for (String path : binaryTreePaths(root.left)) {
paths.add(root.val + "->" + path);
}
for (String path : binaryTreePaths(root.right)) {
paths.add(root.val + "->" + path);
}
return paths;
}
https://xuezhashuati.blogspot.com/2016/02/binary-tree-paths.html
https://zhengyang2015.gitbooks.io/lintcode/content/binary_tree_paths_480.html
public List<String> binaryTreePaths(TreeNode root) {
// Write your code here
List<String> result = new ArrayList<String>();
if(root == null){
return result;
}
if(root.left == null && root.right == null){
result.add(String.valueOf(root.val));
}
List<String> left = binaryTreePaths(root.left);
List<String> right = binaryTreePaths(root.right);
if(left.size() != 0){
for(String s : left){
StringBuilder sb = new StringBuilder();
sb.append(root.val);
sb.append("->");
sb.append(s);
result.add(sb.toString());
}
}
if(right.size() != 0){
for(String s : right){
StringBuilder sb = new StringBuilder();
sb.append(root.val);
sb.append("->");
sb.append(s);
result.add(sb.toString());
}
}
return result;
}