@@ -51,30 +51,39 @@ public List<Integer> postorderTraversal2(TreeNode root) {
5151 }
5252 return ans ;
5353 }
54+ 5455 // TODO
5556 // 【非递归】【单栈】后序遍历
5657 public static List <Integer > postorderTraversal1 (TreeNode root ) {
57- List <Integer > ans = new ArrayList <>();
58- if (null == root ) {
59- return ans ;
58+ List <Integer > result = new ArrayList <>();
59+ if (root == null ) {
60+ return result ;
6061 }
62+ 6163 Stack <TreeNode > stack = new Stack <>();
62- TreeNode c ;
63- TreeNode h = root ;
64- stack .push (h );
65- while (!stack .isEmpty ()) {
66- c = stack .peek ();
67- // 如果c的左孩子和有孩子都不为空,且上一个节点不是c的右孩子,说明是c是新加入的节点,把左孩子压栈
68- if (c .left != null && h != c .left && h != c .right ) {
69- stack .push (c .left );
70- } else if (c .right != null && h != c .right ) {
71- stack .push (c .right );
72- } else {
73- ans .add (stack .pop ().val );
74- h = c ;
64+ TreeNode current = root ; // 当前探索指针
65+ TreeNode lastVisit = null ; // 记录上一个被访问的节点
66+ 67+ while (current != null || !stack .isEmpty ()) {
68+ // 1. 左链入栈:一直向左走,把所有左孩子压栈
69+ while (current != null ) {
70+ stack .push (current );
71+ current = current .left ;
72+ }
73+ 74+ // 2. 查看栈顶节点(不弹出,先判断右子树)
75+ TreeNode peekNode = stack .peek ();
76+ 77+ // 3. 如果右子树存在且未被访问过,则转向右子树
78+ if (peekNode .right != null && peekNode .right != lastVisit ) {
79+ current = peekNode .right ; // 处理右子树
80+ } // 4. 否则(右子树为空或已访问),可以访问当前节点
81+ else {
82+ result .add (peekNode .val ); // 访问节点
83+ lastVisit = stack .pop (); // 记录已访问
7584 }
7685 }
77- return ans ;
86+ return result ;
7887 }
7988
8089 // morris遍历实现后序遍历
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