Leetcode 156. Binary Tree Upside Down

方法1: bottomUp-dfs-postorder-recursion。时间复杂n，空间复杂h。这道题唯一需要知道的点是：先处理左右子树，然后经过处理的左子树的rightmost node左边连经过处理的右子树，右边连一个值为root.val的节点。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    
    public TreeNode upsideDownBinaryTree(TreeNode root) {
    
    
        if(root == null) return root;
        if(root.left == null && root.right == null) return root;
        TreeNode left = upsideDownBinaryTree(root.left);
        TreeNode right = upsideDownBinaryTree(root.right);
        TreeNode copy = left;
        while(copy.right != null){
    
    
            copy = copy.right;
        }
        copy.left = right;
        copy.right = new TreeNode(root.val);
        return left;
    }
}

总结：

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