leetcode 156: Binary Tree Upside Down

标签：

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

    1
   /   2   3
 / 4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4
  /  5   2
    /    3   1  

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

[分析]

搞清此树的定义,答案自明.

[注意]

None

[CODE]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode UpsideDownBinaryTree(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        if(root==null) return null;
        while(root.left != null) {stack.push(root); root=root.left;}
        stack.push(root);
        while(!stack.empty()) {
            TreeNode node = stack.pop();
            if(!stack.empty()) {
                node.right = stack.peek();
                node.left = stack.peek().right;
            } else {
                node.left = null;
                node.right = null;
            }
        }
        return root;
    }
}

leetcode 156: Binary Tree Upside Down

标签：

原文地址：http://blog.csdn.net/xudli/article/details/42362441