力扣labuladong——一刷day55

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前言

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二叉树的遍历代码是动态规划和回溯算法的祖宗。 动态规划 的关键在于明确递归函数的定义，把用子问题的结果推导出大问题的结果。 回溯算法 就简单粗暴多了，就是单纯的遍历回溯树。

一、力扣951. 翻转等价二叉树

/**
 * 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 boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if(root1 == null && root2 == null){
            return true;
        }
        if(root1 == null || root2 == null){
            return false;
        }
        if(root1.val != root2.val){
            return false;
        }
        return (flipEquiv(root1.left,root2.left) && flipEquiv(root1.right,root2.right)) || (
            flipEquiv(root1.left,root2.right) && flipEquiv(root1.right,root2.left)
        );
    }
}

二、力扣124. 二叉树中的最大路径和

/**
 * 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 = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        fun(root);
        return res;
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int l = Math.max(0,fun(root.left));
        int r = Math.max(0,fun(root.right));
        res = Math.max(res,l+r+root.val);
        return Math.max(l,r) + root.val;
    }
}

三、力扣112. 路径总和（遍历）

/**
 * 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 {
    boolean flag = false;
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null){
            return false;
        }
        fun(root,targetSum,0);
        return flag;
    }
    public void fun(TreeNode root, int targetSum, int path){
        if(root == null){
            return;
        }
        if(root.left == null && root.right == null){
            if(path + root.val == targetSum){
                flag = true;
            }
            return;
        }
        fun(root.left,targetSum,path+root.val);
        fun(root.right,targetSum,path+root.val);
    }
}

四、力扣112. 路径总和（分解）

/**
 * 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 boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null){
            return false;
        }
        if(root.left == root.right && root.val == targetSum){
            return true;
        }
        return hasPathSum(root.left,targetSum - root.val) || hasPathSum(root.right,targetSum-root.val);
    }
}