leetcode 114. Flatten Binary Tree to Linked List

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