二叉树非递归遍历

1. 堆栈实现前序遍历

/**
 * 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:
	vector preorderTraversal(TreeNode* root) {
		if (root == nullptr) return vector(0);

		vector ans(0);
		stack st;
		while (root || !st.empty())
		{
			while (root)
			{
				ans.push_back(root->val);
				st.push(root);
				root = root->left;
			}

			TreeNode *node = st.top();
			st.pop();
			
			root = node->right;
		}

		return ans;
	}
};

2. 堆栈实现中序遍历 

/**
 * 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:
    vector inorderTraversal(TreeNode* root) {
        if(root==nullptr) return vector(0);
        
        vector ans(0);
        stack st;
        while(root || !st.empty())
        {
            while(root)
            {
                st.push(root);
                root = root->left;
            }
            
            TreeNode *node = st.top();
            ans.push_back(node->val);
            st.pop();
            
            root = node->right;
        }
        
        return ans;
    }
};

3. 堆栈实现后序遍历

class Solution {
public:
	vector postorderTraversal(TreeNode* root) {
		if (root == nullptr) return vector(0);

		vector ans(0);
		stack st;
		TreeNode *lastViewed = nullptr; //记录上次输出的节点

		//跳到最下边的左子树
		while (root)
		{
			st.push(root);
			root = root->left;
		}
		while (!st.empty())
		{
			//右子树为空或者上次访问过右子树时，输出此节点
			if (st.top()->right == nullptr || st.top()->right == lastViewed)
			{
				ans.push_back(st.top()->val); //输出节点
				lastViewed = st.top(); //这里对lastViewed赋值才有意义，因为这时才算真正输出了节点（即把节点放到ans里）
				st.pop();
			}
			else
			{
				root = st.top()->right;//此时右子树不为空
				while (root)
				{
					st.push(root);
					root = root->left;
				}
			}
		}

		return ans;
	}
};

参考文献：https://blog.csdn.net/zhangxiangDavaid/article/details/37115355