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前面几种遍历方法比较简单，说明一下最后一种
1. 如果路径上前一个节点是父节点，则往左孩子方向走
2. 如果路径上前一个节点是左孩子，则往右孩子方向走
3. 如果路径上前一个节点是右孩子，则往父节点方向走
处理下孩子缺失的情况
1. 如果只有一个右孩子且从父节点过来，往右孩子方向走 （有右无左）
2. 如果从左孩子过来且无右孩子，则往父节点方向走 （有左无右）
3. 如果没有孩子节点，则往父节点走 （无左无右）
#include <iostream> #include <stack> using namespace std; struct Tree{ Tree *parent; Tree *left; Tree *right; int key; }; //递归方式 先序 中序 后序 //先序 void Ergodic_P(Tree *tree){ if (!tree) return; cout<<tree->key; Ergodic_P(tree->left); Ergodic_P(tree->right); } //中序 void Ergodic_M(Tree *tree){ if (!tree) return; Ergodic_M(tree->left); cout << tree->key; Ergodic_M(tree->right); } //后序 void Ergodic_S(Tree *tree){ if (!tree) return; Ergodic_S(tree->left); Ergodic_S(tree->right); cout << tree->key; } //非递归方式 //借用栈 void Ergodic_Stack(Tree *tree){ stack<Tree *> sta; sta.push(tree); while (!sta.empty()){ tree = sta.top(); sta.pop(); cout << tree->key; if (tree->left) sta.push(tree->left); if (tree->right) sta.push(tree->right); } } //非常巧妙的一种方法 void Ergodic_Tricky(Tree *tree){ Tree *prev = 0; while (tree) { if (prev == tree->parent) { cout << tree->key; prev = tree; tree = tree->left ? tree->left : tree->right ? tree->right : tree->parent; } else if (prev == tree->left && tree->right) { prev = tree; tree = tree->right; } else { prev = tree; tree = tree->parent; } } } int main(){ Tree leaf_1{0,0,0,1}; Tree leaf_2{ 0, 0, 0, 2 }; Tree leaf_3{ 0, 0, 0, 3 }; Tree leaf_4{ 0, 0, 0, 4 }; Tree tree_1{ 0, &leaf_1, &leaf_2, 5 }; Tree tree_2{ 0, &leaf_3, &leaf_4, 6 }; Tree root{ 0, &tree_1, &tree_2, 7 }; leaf_1.parent = &tree_1; leaf_2.parent = &tree_1; leaf_3.parent = &tree_2; leaf_4.parent = &tree_2; tree_1.parent = &root; tree_2.parent = &root; Ergodic_P(&root); cout << endl; Ergodic_P(&root); cout << endl; Ergodic_S(&root); cout << endl; Ergodic_Stack(&root); cout << endl; Ergodic_Tricky(&root); }
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原文地址：http://www.cnblogs.com/Nastukashii/p/4420497.html
