非递归遍历树的总结（前中后序）

递归，显而易见。

非递归，bfs，是用队列。

而前中后序，一般都是用栈。

一种很直观的方法是，采用一个flag，记录第几次进栈出栈，从而决定是否打印，决定下一个入栈的是左还是右子节点。

 

但是下面的方法，更近精简和巧妙。

https://discuss.leetcode.com/topic/30632/preorder-inorder-and-postorder-iteratively-summarization

把前、中、后三种遍历都总结了，而且没有用多余的flag。采用的技巧是同时判断栈和cur节点。

前序：

public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.add(p.val);  // Add before going to children
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            p = node.right;   
        }
    }
    return result;
}

View Code

中序:

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            result.add(node.val);  // Add after all left children
            p = node.right;   
        }
    }
    return result;
}

View Code

后序：

public List<Integer> postorderTraversal(TreeNode root) {
    LinkedList<Integer> result = new LinkedList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.addFirst(p.val);  // Reverse the process of preorder
            p = p.right;             // Reverse the process of preorder
        } else {
            TreeNode node = stack.pop();
            p = node.left;           // Reverse the process of preorder
        }
    }
    return result;
}

View Code

 

另外，对于后序，有一种解法，是先打印，然后再反转。

https://discuss.leetcode.com/topic/2919/my-accepted-code-with-explaination-does-anyone-have-a-better-idea

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        stack<TreeNode*> nodeStack;
        vector<int> result;
        //base case
        if(root==NULL)
        return result;
        nodeStack.push(root);
    while(!nodeStack.empty())
    {
        TreeNode* node= nodeStack.top();  
        result.push_back(node->val);
        nodeStack.pop();
        if(node->left)
        nodeStack.push(node->left);
        if(node->right)
        nodeStack.push(node->right);
    }
     reverse(result.begin(),result.end());
     return result;

}

View Code

 

也很巧妙。

 

上面这个解法，也是针对这道题目：

https://leetcode.com/problems/binary-tree-postorder-traversal/?tab=Description