Invert Binary Tree

nvert a binary tree.

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

to

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

这道题也称为求二叉树的镜像

TreeNode* invertTree(TreeNode* root) {
        if(root == NULL) return root;

        TreeNode *newLeft = invertTree(root->right);
        TreeNode *newRight = invertTree(root->left);

        root->left = newLeft;
        root->right = newRight;

        return root;
    }

/**
 * 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* r) {
        if(r == NULL)   return r;

        if(r->left == NULL && r->right == NULL) return r;

        swapTreeChild(r);

        if(r->left)  invertTree(r->left);

        if(r->right)    invertTree(r->right);

        return r;
    }
private:
    void swapTreeChild(TreeNode *p){
        TreeNode *temp = p->left;
        p->left = p->right;
        p->right = temp;
    }
};

迭代的版本

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == NULL)    return root;

        if(root->left == NULL && root->right == NULL)   return root;

        stack<TreeNode *>st;
        st.push(root);

        while(!st.empty()){
            TreeNode *temp = st.top();
            st.pop();
            if(temp->left) st.push(temp->left);
            if(temp->right) st.push(temp->right);
            swapTreeChild(temp);
        }
        return root;
    }
private:
    void swapTreeChild(TreeNode *p){
        TreeNode *temp = p->left;
        p->left = p->right;
        p->right = temp;
    }
};