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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.
For example:
Given a binary tree {1,2,3,4,5},
1 / 2 3 / 4 5
return the root of the binary tree [4,5,2,#,#,3,1].
4
/ 5 2
/ 3 1
根据题给定的tree的特性，新生产的tree的所有左子树均为原树的右子树。那么我们只有一个stack来存左子树就好了，然后pop第一个值作为新树的root，stack的下一个设为a为root的右子树，如果a有右子树，则root左子树为a的右子树，依次类推直到stack为空。
注意要讲新树的左子树设为null，t.right.left =null; 不然会出现无限循环。
public TreeNode UpsideDownBinaryTree(TreeNode root) { if(root == null) return null; var stack = new Stack<TreeNode>(); while(root != null) { stack.Push(root); root = root.left; } var sentinel = new TreeNode(-1); var t = stack.Pop(); sentinel.left = t; while(stack.Count()>0) { var next = stack.Pop(); if(next.right != null) { t.left = next.right; next.right = null; } t.right =next; t.right.left =null; t = t.right; } return sentinel.left; }
迭代的方法为
public TreeNode UpsideDownBinaryTree(TreeNode root) { return DFS(root); } private TreeNode DFS(TreeNode root) { if(root == null || root.left == null) { return root; } else { var l = root.left; var r = root.right; var res = DFS(root.left); l.right =root; l.left = r; root.left = null;root.right = null; return res; } }
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原文地址：http://www.cnblogs.com/renyualbert/p/5904290.html
