Leetcode 226. Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null){
            return root;
        }
        if(root.left == null && root.right == null){
            return root;
        }
        if(root.left == null){
            root.left = root.right;
            root.right = null;
            invertTree(root.left);
        }
        else if(root.right == null){
            root.right = root.left;
            root.left = null;
            invertTree(root.right);
        }
        else{
            TreeNode tempnode = root.left;
            root.left = root.right;
            root.right = tempnode;
            invertTree(root.left);
            invertTree(root.right);
        }
        return root;
    }
}

犯错：

else if(root.right == null)之前是 if(root.right == null)，造成刚刚翻转过来的树又翻转回去

教训：

if

else if

else是非此即彼的选项

而if

if是验证两遍

discuss中更简单的解法，还需再研究看懂

<span style="font-family: Arial, Helvetica, sans-serif;">public class Solution {</span><p class="p1"><span class="s1">    public TreeNode invertTree(TreeNode root) {</span></p><p class="p1"><span class="s1">        if (root == null)</span></p><p class="p1"><span class="s1">            return root;</span></p><p class="p2"><span class="s1"></span>
</p><p class="p1"><span class="s1">        TreeNode tmp = root.left;</span></p><p class="p1"><span class="s1">        root.left = invertTree(root.right);</span></p><p class="p1"><span class="s1">        root.right = invertTree(tmp);</span></p><p class="p2"><span class="s1"></span>
</p><p class="p1"><span class="s1">        return root;</span></p><p class="p1"><span class="s1">    }</span></p><p class="p1"><span class="s1">}</span></p>