LeetCode Invert Binary Tree
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Invert a binary tree.
4 / \ 2 7 / \ / \1 3 6 9to
4 / \ 7 2 / \ / \9 6 3 1反转二叉树,首先采用递归的方式。先遍历作左子树,当指针指向NULL时,返回,赋值给tmp。接着遍历右子树,指向NULL,返回,并将该返回值赋值给左孩子,然后将tmp赋值给右孩子。此时完成了节点1的两个左右节点的交换,然后返回上一层递归,tmp指向1节点,然后遍历右子树,重复第一步步骤,完成反转。接下来的每个节点按照如上操作。知道所有节点完成反转。
/** * 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) { if(root == NULL){ return root; } TreeNode *tmp = invertTree(root->left); root->left = invertTree(root->right); root->right = tmp; return root;};第二种方法采用非递归的形式完成,此时需要借助一个stack<TreeNode*>存储遍历过的根节点。从root先反转左右子树,root入栈,然后走向左子树,此时左子树为7,而不是原来的2。接着交换根节点为7的左右子树,7入栈。当左孩子为NULL时(此时左孩子的父节点已完两个左右子女的交换),父节点出栈,指针指向右子树。重复上述过程。这个交换过程很类似于先序遍历,不过每次根节点的两个子女要进行交换。
class Solution {public: TreeNode* invertTree(TreeNode* root) { stack<TreeNode*> s; TreeNode* p = root; TreeNode* tmp = NULL; while (p || !s.empty()) { while (p) { tmp = p -> left; p -> left = p -> right; p -> right = tmp; s.push(p); p = p -> left; } p = s.top(); s.pop(); p = p -> right; } return root; }};
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