• 二叉树的前中后序遍历(非递归)


    前序遍历一:

        vector<int> preorderTraversal(TreeNode* root) {
            stack<TreeNode*> S;
            vector<int> v;
            TreeNode* rt = root;
            while(rt || S.size()){
                while(rt){
                    S.push(rt->right);
                    v.push_back(rt->val);
                    rt=rt->left;
                }
                rt=S.top();S.pop();
            }
            return v;        
        }

    前序遍历二:

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if(!root) return res;
        TreeNode* p = root;
        stack<TreeNode*> S;
        S.push(root);
        while(!S.empty()) {
            p = S.top();
            S.pop();
            res.push_back(p->val);
            if(p->right) S.push(p->right);
            if(p->left) S.push(p->left);
        }
        return res;
    }

    中序遍历:

    vector<int> inorderTraversal(TreeNode* root) 
    {
        stack<TreeNode*> S;
        vector<int> v;
        TreeNode* rt = root;
        while(rt || S.size()){
            while(rt){
                S.push(rt);
                rt=rt->left;
            }
            rt=S.top();S.pop();
            v.push_back(rt->val);
            rt=rt->right;
        }
        return v;  
    }

    后序遍历一:

        vector<int> postorderTraversal(TreeNode* root) {
            stack<TreeNode*> S;
            vector<int> v;
            TreeNode* rt = root;
            while(rt || S.size()){
                while(rt){
                    S.push(rt->left);
                    v.push_back(rt->val);
                    rt=rt->right;
                }
                rt=S.top();S.pop();
            }
            reverse(v.begin(),v.end());
            return v;
        }

    后序遍历二:

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> S;
        TreeNode *rt = root, *pre = NULL;
        while (rt || !S.empty()) {
            while (rt) {//走到最左边
                S.push(rt);
                rt = rt->left;
            }
            
            rt = S.top();
            if (rt->right && rt->right != pre)//右子树存在,未被访问
                rt = rt->right;
            else {
                S.pop();
                res.push_back(rt->val);
                pre = rt;    //记录最近访问过的节点
                rt = NULL;    //回溯的重要步骤
            }
        }
        return res;
    }

    后序遍历三:

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> S;
        TreeNode *rt = root;
        TreeNode *pre = NULL;
        if(rt == NULL) return res;
        S.push(root);
        while(!S.empty()) {
            rt = S.top();
            if(!rt->left && !rt->right || (pre && (pre == rt->left || pre == rt->right))) {
                res.push_back(rt->val);
                S.pop();
                pre = rt;
            }
            else {
                if(rt->right) S.push(rt->right);
                if(rt->left) S.push(rt->left);
            }
        }
        return res;
    }

    后序遍历四:

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> S;
        if(root) S.push(root);
        while(!S.empty()) {
            TreeNode *t = S.top();
            S.pop();
            if(t) {
                S.push(t);         //在右节点之前重新插入该节点,以便在最后处理(访问值)
                S.push(nullptr);   //nullptr跟随t插入,标识已经访问过,还没有被处理
                if(t->right) S.push(t->right);
                if(t->left) S.push(t->left);
            } else {
                res.push_back(S.top()->val);
                S.pop();
            }
        }
        return res;   
    }
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  • 原文地址:https://www.cnblogs.com/joker1937/p/12857472.html
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