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1. 蠢蠢的方法,dfs保存,重新构造
1 public void flatten(TreeNode root) { 2 if(root == null) { 3 return; 4 } 5 Stack<TreeNode> stack = new Stack<TreeNode>(); 6 LinkedList<TreeNode> nodes = new LinkedList<TreeNode>(); 7 stack.push(root); 8 while(!stack.isEmpty()) { 9 TreeNode cur = stack.pop(); 10 nodes.add(cur); 11 if(cur.right != null) { 12 stack.push(cur.right); 13 } 14 if(cur.left != null) { 15 stack.push(cur.left); 16 } 17 18 } 19 TreeNode pre = nodes.poll(); 20 while(!nodes.isEmpty()) { 21 pre.left = null; 22 TreeNode cur = nodes.poll(); 23 pre.right = cur; 24 pre = cur; 25 } 26 }
2. 用递归,如果右节点不是空的就先保存下来,然后root的右节点等于左节点,左节点清空,cur的节点要么是root的右节点(即原先的左节点),要么就从栈里弹一个出来,递归
1 public void flatten(TreeNode root) { 2 if(root == null) { 3 return; 4 } 5 Stack<TreeNode> stack = new Stack<TreeNode>(); 6 helper(root, stack); 7 } 8 9 private void helper(TreeNode root, Stack<TreeNode> stack) { 10 if(root == null) { 11 return; 12 } 13 TreeNode right = root.right; 14 if(right != null) { 15 stack.push(right); 16 } 17 TreeNode cur = null; 18 if(root.left == null) { 19 if(!stack.isEmpty()) { 20 cur = stack.pop(); 21 } else { 22 return; 23 } 24 } else { 25 cur = root.left; 26 } 27 root.right = cur; 28 root.left = null; 29 helper(cur, stack); 30 }
3. code ganker
ref:http://blog.csdn.net/linhuanmars/article/details/23717703
1 public void flatten(TreeNode root) { 2 ArrayList<TreeNode> pre = new ArrayList<TreeNode>(); 3 pre.add(null); 4 helper(root, pre); 5 } 6 private void helper(TreeNode root, ArrayList<TreeNode> pre) 7 { 8 if(root == null) 9 return; 10 TreeNode right = root.right; 11 if(pre.get(0)!=null) 12 { 13 pre.get(0).left = null; 14 pre.get(0).right = root; 15 } 16 pre.set(0,root); 17 helper(root.left, pre); 18 helper(right, pre); 19 }
114. Flatten Binary Tree to Linked List
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原文地址:http://www.cnblogs.com/warmland/p/5582636.html