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思路

反着的中序遍历，并计数

代码

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {//遍历次数int c=0;//要找的数字int cnt;int result;boolean enough=false;//寻找第cnt大的数public int findTargetNode(TreeNode root, int cnt) {this.cnt=cnt;fMidOrder(root);return result;}/*** 反中序遍历，右中左，遍历一次累计一次，当次数等于cnt,result就是这个数，* @param root* @return*/public void fMidOrder(TreeNode root) {//判断是否足够if (enough||root==null) {return;}//遍历右边if (root.right!= null) {fMidOrder(root.right);}//遍历自己c++;if (c==cnt ) {result=root.val;enough=true;return;}//遍历左边if (root.left!= null) {fMidOrder(root.left);}}}

记录

总结