LeetCode145 Binary Tree Postorder Traversal 二叉树后序遍历

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

题源：here；完整实现：here

后序遍历就是：左->右->root的遍历方式（因为root在后面），同样有两种解决方案，但是请注意，后序遍历的难度是hard，所以在用非递归的方式实现的过程也更复杂一些。

1 递归的实现方案

	vector<int> postorderTraversal(TreeNode* root) {
		vector<int> res;
		if (!root) return res;
		vector<int> left;
		vector<int> right;

		if (root->left) left = postorderTraversal(root->left);
		if (root->right) right = postorderTraversal(root->right);
		
		if (!left.empty()) {
			res = left;
		}
		if (!right.empty()) {
			res.insert(res.end(), right.begin(), right.end());
		}
		res.push_back(root->val);

		return res;
	}

2 利用栈完成遍历

	vector<int> postorderTraversal2(TreeNode* root) {
		vector<int> res;
		if (!root) return res;
		stack<TreeNode*> s;
		TreeNode* lastV = NULL;
		TreeNode* tmp = root;

		while (!s.empty() || tmp) {
			if (tmp) {
				s.push(tmp);
				tmp = tmp->left;
			}
			else {
				TreeNode* n = s.top();
				if (n->right && lastV != n->right) {
					tmp = n->right;
				}
				else {
					res.push_back(n->val);
					s.pop();
					lastV = n;
				}
			}
		}

		return res;
	}