Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / 2 5 / 3 4 6
The flattened tree should look like:
1 2 3 4 5 6
Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
思路:
对整颗树向右子树方向遍历,
如果遍历的当前节点有右子树,将这个右侧子节点入栈,
如果有左子树就将左子树放在右边,左子树置为空,
如果没有左子树说明这个点是某个左子树的最后一个左侧子节点,如果此时栈不为空,将栈内的最后一个节点拿出来作为节点的右子树。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public void flatten(TreeNode root) { Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode node = root; while (node != null || !stack.isEmpty()) { if (node.right != null) { stack.push(node.right); } if (node.left != null) { node.right = node.left; node.left = null; }else if (!stack.isEmpty()){ TreeNode temp = stack.pop(); node.right = temp; } node = node.right; } } }
先把root存起来,(存到node节点)
node、stack非空进循环
右不为空右压栈
左不为空,右等于左,左置空,
左空栈不空,弹栈赋给右,
node右移出循环