leetcode257 Binary Tree Paths

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Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1 /   \2     3 \  5

All root-to-leaf paths are:

["1->2->5", "1->3"]import java.util.ArrayList;import java.util.List;/** * Given a binary tree, return all root-to-leaf paths.   For example, given the following binary tree:   1 /   \2     3 \  5  All root-to-leaf paths are:["1->2->5", "1->3"] * * */public class Solution {<span style="white-space:pre"></span>/*树的后序遍历代码*/public List<String> binaryTreePaths1(TreeNode root) {ArrayList<TreeNode> treeList = new ArrayList<TreeNode>();List<String> res = new ArrayList<String>();TreeNode p = null;int flag, i, top = -1;if (root != null) {do {while (root != null) {top++;treeList.add(root);root = root.left;}p = null;flag = 1;while (top != -1 && flag != 0) {root = treeList.get(top);if (root.right == p) {/* * if(root.left==null && root.right==null) { * System.out.println(root.val); } */System.out.println(root.val);treeList.remove(top);top--;p = root;} else {root = root.right;flag = 0;}}} while (top != -1);}return res;}<span style="white-space:pre"></span>/*可以AC的程序代码*/public List<String> binaryTreePaths(TreeNode root) {ArrayList<TreeNode> treeList = new ArrayList<TreeNode>();List<String> res = new ArrayList<String>();int flag, i, top = -1;TreeNode q = root;TreeNode p = null;do {while (q != null) {top++;treeList.add(q);q = q.left;}p = null;flag = 1;while (top != -1 && flag != 0) {q = treeList.get(top);if (q.right == p) {//System.out.println(q.val);if(q.left==null && q.right ==null){String re = "";for(int i1 = 0;i1<top;i1++){re+=(treeList.get(i1).val+"->");}re+=treeList.get(top).val;res.add(re);}treeList.remove(top);top--;p = q;} else {q = q.right;flag = 0;}}} while (top != -1);/*for(int i1 = 0;i1<res.size();i1++){System.out.println(res.get(i1));}*/return res;}public static void main(String[] args) {TreeNode root = new TreeNode(1);TreeNode l1 = new TreeNode(2);TreeNode l2 = new TreeNode(3);TreeNode l3 = new TreeNode(5);root.left = l1;root.right = l2;l1.right = l3;new Solution().binaryTreePaths(root);}}class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}

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