leetcode 226. Invert Binary Tree

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Invert a binary tree.

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

     4
   /   \
  7     2
 / \   / \
9   6 3   1
二叉树的递归实现：

<span style="font-size:18px;">/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ struct TreeNode* invertTree(struct TreeNode* root) { if(root==NULL) return root; invertTree(root->left); invertTree(root->right); struct TreeNode* T; T=root->left; root->left=root->right; root->right=T; return root; }</span> <span style="font-size:18px;">然后是非递归形式的，使用队列实现,： </span><pre name="code" class="cpp"><span style="font-size:18px;">/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* invertTree(TreeNode* root) { queue<TreeNode*> q; if(root) q.push(root); struct TreeNode *c,*tmp; while(!q.empty()) { c=q.front(); q.pop(); tmp=c->left; c->left=c->right; c->right=tmp; if(c->left) q.push(c->left); if(c->right) q.push(c->right); } return root; } };</span>

好久没有做过题啦，写bfs和dfs都感觉犹豫：-（