题解 | #二叉搜索树的第k个节点#

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param proot TreeNode类 
     * @param k int整型 
     * @return int整型
     */
    public int KthNode (TreeNode proot, int k) {
        // write code here
        if(proot == null || k < 0){
            return -1;
        }
        //中序 遍历 左 中 右
        Stack<TreeNode> stack = new Stack<>();
        stack.push(proot);
        TreeNode node = proot;
        int i = 0;
        while(!stack.isEmpty()){
            while(node.left != null){
                stack.push(node.left);
                node = node.left;
            }
            i++;
            TreeNode temp = stack.pop();
            if(i == k){
                return temp.val;
            }
            if(temp.right != null){
                stack.push(temp.right);
                node = temp.right;
            }
            
        }
        return -1;
    }
}