最优雅的二叉树非递归遍历模板

时间 2020-01-23

原文原文链接

先压栈的是右孩子，再是左孩子，因为栈先进后出的特色，先取出来的就是左孩子，而后是右孩子，知足 根 - 左 - 右。post

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return res;
        stack<TreeNode*> s;
        s.push(root);
        TreeNode *node;
        while (!s.empty()) {
            node = s.top();
            s.pop();
            res.push_back(node->val);
            if (node->right) s.push(node->right);
            if (node->left) s.push(node->left);
        }
        return res;
    }
};

LeetCode 94. 二叉树的中序遍历code

直接模拟中序遍历的过程。leetcode

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return res;
        TreeNode *node = root;
        stack<TreeNode*> s;
        while (!s.empty() || node) {
            if (node) {
                s.push(node);
                node = node->left;
            } else {
                node = s.top();
                s.pop();
                res.push_back(node->val);
                node = node->right;
            }
        }
        return res;
    }
};

LeetCode 145. 二叉树的后序遍历get

有点抽象，总之只需理解 根 - 右 - 左 逆过来就是 左 - 右 - 根。io

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