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145. Binary Tree Postorder Traversal QuestionEditorial Solution

Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1         2    /   3

 

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

 

每次都先push right, 然后left。 但是需要一个sentinel存每次上一次pop的node, 如果这个node是新的peek的值的子数的话,不能再push到stack,需要进入输出逻辑。 同样进入输出逻辑的还有leaf。

    public IList<int> PostorderTraversal(TreeNode root) {        var res = new List<int>();        var stack = new Stack<TreeNode>();        if(root == null) return res;        stack.Push(root);        var rightPointer = new TreeNode(-1);        while(stack.Count() > 0)        {            if((root.right == null && root.left == null)||(root.left == rightPointer ) || (root.right == rightPointer) )            {                res.Add(root.val);                rightPointer = stack.Pop();                if(stack.Count()>0)                root = stack.Peek();            }            else            {                if(root.right != null) stack.Push(root.right);                if(root.left != null) stack.Push(root.left);                root = stack.Peek();            }                    }        return res;    }

 

145. Binary Tree Postorder Traversal QuestionEditorial Solution