Flatten Binary Tree to Linked List leetcode

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

click to show hints.

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