[Algorithms] 二叉树的迭代遍历

2025-04-24 19:09:06   2025-04-24 19:09:06
Categories Algorithms BinaryTree
Tags

414 Words   |   2 Minutes

来源：代码随想录

LeetCode 144. 二叉树的前序遍历

LeetCode 145. 二叉树的后序遍历

LeetCode 94. 二叉树的中序遍历

掌握一种迭代方法（非统一/统一）即可。这里只掌握非统一迭代。

前序

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode *> st;
        vector<int> v;
        if (root != nullptr) {
            st.push(root);
        }
        while (!st.empty()) {
            TreeNode *node = st.top();
            st.pop();
            v.push_back(node->val);
            if (node->right) {
                st.push(node->right);
            }
            if (node->left) {
                st.push(node->left);
            }
        }

        return v;
    }
};

---

中序

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode *> st;
        vector<int> v;
        TreeNode *cur = root;
        while (!st.empty() || cur != nullptr) {
            if (cur != nullptr) {
                st.push(cur);
                cur = cur->left;
            } else {
                cur = st.top();
                st.pop();
                v.push_back(cur->val);
                cur = cur->right;
            }
        }

        return v;
    }
};

---

后序

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode *> st;
        vector<int> v;
        if (root != nullptr) {
            st.push(root);
        }
        while (!st.empty()) {
            TreeNode *node = st.top();
            st.pop();
            v.push_back(node->val);
            if (node->left) {
                st.push(node->left);
            }
            if (node->right) {
                st.push(node->right);
            }
        }

        reverse(v.begin(), v.end());
        return v;
    }
};

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