力扣labuladong——一刷day59

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前言

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像求和、求高度这种基本的二叉树函数很容易写，有时候只要在它们的后序位置添加一点代码，就能得到我们想要的答案。

一、力扣549. 二叉树中最长的连续序列

/**
 * 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 {
    int res = 0;
    public int longestConsecutive(TreeNode root) {
        fun(root);
        return res;
    }
    public int[] fun(TreeNode root){
        if(root == null){
            return new int[]{0,0};
        }
        int[] left = fun(root.left);
        int[] right = fun(root.right);
        int leftInc = left[0], leftDes = left[1];
        int rightInc = right[0], rightDes = right[1];
        int curInc = 1, curDes = 1;
        if(root.left != null){
            if(root.left.val -1 == root.val){
                curInc += leftInc;
            }
            if(root.left.val + 1 == root.val){
                curDes += leftDes;
            }
        }
        if(root.right != null){
            if(root.right.val - 1 == root.val){
                curInc = Math.max(curInc,rightInc+1);
            }
            if(root.right.val + 1 == root.val){
                curDes = Math.max(curDes, rightDes+1);
            }
        }
        res = Math.max(res, curInc + curDes -1);
        return new int[]{curInc,curDes};
    }
}

二、力扣1325. 删除给定值的叶子节点

/**
 * 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 {
    public TreeNode removeLeafNodes(TreeNode root, int target) {
        if(root == null){
            return null;
        }
        if(root.left == null && root.right == null){
            if(root.val == target){
                return null;
            }
        }
        TreeNode left = removeLeafNodes(root.left,target);
        TreeNode right = removeLeafNodes(root.right,target);
        if(left == null && right == null){
            if(root.val == target){
                return null;
            }
        }
        root.left = left;
        root.right = right;
        return root;
    }
}